Solve the quadratic equation (2x+3)(x-3)=7. If there is more than one correct answer, enter your answers as comma separated list. If there are no real solutions,enter none. Please help!!!!! :(
It is good to have the quadratic in factored form but useful only if the product is zero.
In your case it is 7, so that is useless.
Expand and recalculate the constant ....
2x^2 - 6x + 3x - 9 = 7
2x^2 - 3x - 16 =0
hint: it does not factor , so you have to use the quadratic formula
To solve the quadratic equation (2x+3)(x-3)=7, we need to simplify the equation first.
Step 1: Expand the left side of the equation:
(2x + 3)(x - 3) = 7
2x^2 - 6x + 3x - 9 = 7
2x^2 - 3x - 9 = 7
Step 2: Move all terms to one side of the equation:
2x^2 - 3x - 9 - 7 = 0
2x^2 - 3x - 16 = 0
Now that we have the equation in the standard quadratic form of ax^2 + bx + c = 0, we can solve it using the quadratic formula. The quadratic formula is given as:
x = (-b ± √(b^2 - 4ac)) / (2a)
Comparing the equation to the standard format, we have:
a = 2, b = -3, and c = -16.
Step 3: Substitute the values into the quadratic formula and solve for x:
x = (-(-3) ± √((-3)^2 - 4(2)(-16))) / (2(2))
x = (3 ± √(9 + 128)) / 4
x = (3 ± √137) / 4
The two solutions for x are:
x = (3 + √137) / 4
x = (3 - √137) / 4
Therefore, the solutions to the quadratic equation (2x+3)(x-3)=7 are (3 + √137) / 4 and (3 - √137) / 4.