r^2=3r+4

Solution for r^2=3r+4 equation:

r^2=3r+4

We move all terms to the left:

r^2-(3r+4)=0

We get rid of parentheses

r^2-3r-4=0

a = 1; b = -3; c = -4;
Δ = b2-4ac
Δ = -32-4·1·(-4)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5}{2*1}=\frac{-2}{2}
=-1
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5}{2*1}=\frac{8}{2}
=4
$