At what values of x does the quadratic function y = 5x ^ 2-4x-4 take on the value 8?

In order to find at what values of the variable x the quadratic function y = 5x ^ 2 – 4x – 4 takes a value equal to 8, we start by substituting it into the equation instead of y = 8.

5x ^ 2 – 4x – 4 = 8;

We solve the resulting quadratic equation:

5x ^ 2 – 4x – 4 – 8 = 0;

5x ^ 2 – 4x – 12 = 0;

Let us first of all calculate the discriminant of the equation:

D = b ^ 2 – 4ac = (-4) ^ 2 – 4 * 5 * (-12) = 16 + 240 = 256;

Let’s calculate the roots of the equation:

x1 = (-b + √D) / 2a = (- (- 4) + √256) / 2 * 5 = (4 + 16) / 10 = 20/10 = 2;

x2 = (-b – √D) / 2a = (- (- 4) – √256) / 2 * 5 = (4 – 16) / 10 = -12/10 = -1.2.