2x squared plus 6 x plus 3 is equal to 0.....
2x^2 + 6x +3 = 0
That does not factor into terms with integers. Use the quadratic formula
x= [-b +/-sqrt(b^2-4ac)]/2a
x = [-6 +/-sqrt(36 -24)]/4
= [-6 +/-sqrt12]/4
= -2.366 or -0.634
2 x ^ 2 + 6 x + 3 = 0
Compare your equation to the standard form :
a x ^ 2 + b x + c = 0
and identify the values for a , b , c.
a = 2
b = 6
c = 3
Now evaluate the discriminant :
b ^ 2 - 4 a c = 6 ^ 2 - 4 * 2 * 3 = 36 -
24 = 12
Since b ^ 2 - 4 ac > 0 there are two unequal real solutions:
x1 = [ - b + sqrt ( b ^ 2 - 4 a c ) ] / 2 a
x1 = [ - 6 + sqrt ( 12 ) ] / 2 * 2
x1 = [ - 6 + sqrt ( 4 * 3 ) ] / 4
x1 = [ - 6 + 2 * sqrt ( 3 ) ] / 4
x1 = 2 * [ - 3 + sqrt ( 3 ) ] / 2 * 2
x1 = [ - 3 + sqrt ( 3 ) ] / 2
x1 = [ sqrt ( 3 ) - 3 ] / 2
x1 = - 0.63 approx.
x2 = [ - b - sqrt ( b ^ 2 - 4 a c ) ] / 2 a
x2 = [ - 6 - sqrt ( 12 ) ] / 2 * 2
x2 = [ - 6 - sqrt ( 4 * 3 ) ] / 4
x2 = [ - 6 - 2 * sqrt ( 3 ) ] / 4
x2 = 2 * [ - 3 - sqrt ( 3 ) ] / 2 * 2
x2 = [ - 3 - sqrt ( 3 ) ] / 2
x2 = [ - 3 - sqrt ( 3 ) ] / 2
x2 = - 2.37 approx.
To solve the quadratic equation 2x^2 + 6x + 3 = 0, we can use the quadratic formula or factorization method. I will explain both methods:
1. Quadratic Formula:
The quadratic formula is given by x = (-b ± √(b^2 - 4ac)) / (2a). Here, a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0.
For our given equation 2x^2 + 6x + 3 = 0, we have:
a = 2, b = 6, c = 3.
Substituting these values into the quadratic formula, we get:
x = (-6 ± √(6^2 - 4*2*3))/(2*2)
= (-6 ± √(36 - 24))/(4)
= (-6 ± √(12))/(4)
Simplifying further, we have:
x = (-6 ± √(12))/(4)
= (-6 ± 2√(3))/(4)
= -3/2 ± (√(3)/2)
Therefore, the solutions to the quadratic equation 2x^2 + 6x + 3 = 0 are:
x = -3/2 + (√(3)/2) and x = -3/2 - (√(3)/2).
2. Factorization Method:
To factorize the equation 2x^2 + 6x + 3 = 0, we need to find two binomials that multiply to give the quadratic equation. The factors will have the form (x + p)(x + q), where p and q are two constants.
We can find the factors by splitting the middle term (6x) into two terms whose coefficients multiply to give ac (2*3 = 6) and whose sum gives the middle term (6x). In this case, the numbers are 3 and 2.
So, we rewrite the equation as follows:
2x^2 + 3x + 2x + 3 = 0
Now, factor by grouping:
(2x^2 + 3x) + (2x + 3) = 0
x(2x + 3) + (2x + 3) = 0
Notice that we have a common factor (2x + 3) in both terms. Factoring it out, we get:
(2x + 3)(x + 1) = 0
Setting each factor equal to zero, we have two possible solutions:
2x + 3 = 0 --> x = -3/2
x + 1 = 0 --> x = -1
Thus, the solutions to the quadratic equation 2x^2 + 6x + 3 = 0 are:
x = -3/2 and x = -1.