Using the quadratic formula, find the two solutions of 4x^2 + 20x + 21 = 0.

Step-by-step solution

1. 1

   Match the equation to the standard form. Comparing $4x^{2}+20x+21=0$ with $ax^{2}+bx+c=0$ gives $a=4$, $b=20$, $c=21$. The leading coefficient is not $1$, so factoring by inspection is awkward and the formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ is the direct route.

2. 2

   Compute the discriminant first. $b^{2}-4ac=20^{2}-4(4)(21)=400-336=64 .$ It is positive, so there are two distinct real roots, and it is a perfect square, so the roots will be rational.

3. 3

   Substitute into the formula. Since $\sqrt{64}=8$ and $2a=8$, $x=\frac{-20\pm 8}{8}.$

4. 4

   Split the plus-or-minus into the two roots. Taking the plus sign, $x=\frac{-20+8}{8}=\frac{-12}{8}=-\frac32$. Taking the minus sign, $x=\frac{-20-8}{8}=\frac{-28}{8}=-\frac72$.

5. 5

   Check both roots by substitution. For $x=-\frac32$: $4\cdot\frac94-30+21=9-30+21=0$. For $x=-\frac72$: $4\cdot\frac{49}{4}-70+21=49-70+21=0$. Both check exactly, and their sum $-5$ matches $-b/a=-20/4$, while their product $\frac{21}{4}$ matches $c/a$.