4r2+3r=12

Solution for 4r2+3r=12 equation:

4r^2+3r=12

We move all terms to the left:

4r^2+3r-(12)=0

a = 4; b = 3; c = -12;
Δ = b2-4ac
Δ = 32-4·4·(-12)
Δ = 201
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{-(3)-\sqrt{201}}{2*4}=\frac{-3-\sqrt{201}}{8}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{201}}{2*4}=\frac{-3+\sqrt{201}}{8}
$