r^2=7r+12=0

Solution for r^2=7r+12=0 equation:

r^2=7r+12=0

We move all terms to the left:

r^2-(7r+12)=0

We get rid of parentheses

r^2-7r-12=0

a = 1; b = -7; c = -12;
Δ = b2-4ac
Δ = -72-4·1·(-12)
Δ = 97
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-\sqrt{97}}{2*1}=\frac{7-\sqrt{97}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{97}}{2*1}=\frac{7+\sqrt{97}}{2}
$