r^2+10r+21=-3

Solution for r^2+10r+21=-3 equation:

r^2+10r+21=-3

We move all terms to the left:

r^2+10r+21-(-3)=0

We add all the numbers together, and all the variables

r^2+10r+24=0

a = 1; b = 10; c = +24;
Δ = b2-4ac
Δ = 102-4·1·24
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2}{2*1}=\frac{-12}{2}
=-6
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2}{2*1}=\frac{-8}{2}
=-4
$