how do i solve for y in the equation -5y(6y - 6) - y = -4(y - 5)
-5y(6y - 6) - y = -4(y - 5)
-30y^2 + 30y - y = -4y + 20
30y^2 - 33y + 20 = 0
y = (22 ± √-1311)/60
or
(22 ± i √1311)/60 , two complex roots.
how kan i solve this equation
7a-3a+2a-a=16
To solve for y in the given equation, let's simplify the equation step by step:
1. Distribute the -5y to the terms inside the parentheses, and distribute the -4 to the term inside the other parentheses:
-5y * 6y + 5y * 6 - y = -4 * y + 4 * 5
This simplifies to:
-30y^2 + 30y - y = -4y + 20
2. Combine like terms:
-30y^2 + 30y - y + 4y = 20
Simplifying further:
-30y^2 + 33y = 20
3. Move all terms to one side of the equation to set it equal to zero:
-30y^2 + 33y - 20 = 0
Now we have a quadratic equation in the form: ax^2 + bx + c = 0, where a = -30, b = 33, and c = -20.
4. To solve the quadratic equation, we can either factor it, complete the square, or use the quadratic formula. In this case, factoring may not be straightforward, so let's use the quadratic formula:
The quadratic formula is given as:
x = (-b ± √(b^2 - 4ac)) / (2a)
Applying this formula to our equation:
y = (-33 ± √(33^2 - 4 * -30 * -20)) / (2 * -30)
Simplifying the equation further:
y = (-33 ± √(1089 - 2400)) / -60
y = (-33 ± √(-1311)) / -60
At this point, we notice that the radicand (number inside the square root) is negative, which means the solutions are imaginary.
Hence, the equation does not have real number solutions for y.