## Precalculus: Mathematics for Calculus, 7th Edition

The real solutions to the equation $4x^{2}-4x-15=0$ are $x=\frac{5}{2}$ and x=$-\frac{3}{2}$.
1. Factor $4x^{2}+4x-15$. The factors are $(-2x-3)(-2x+5)$ because $-2x\times-2x =4x^{2}$(the a value in $a^{2}+bx+c$) and, $-3\times5 = -15$, the c value in the equation. 2. Set the factors equal to zero: $-2x-3=0$ and $-2x+5 = 0$ 3. Solve: $x=-\frac{3}{2}$ and $x=\frac{5}{2}$