3(4r+5)-2r(3r+4)=7

Solution for 3(4r+5)-2r(3r+4)=7 equation:

3(4r+5)-2r(3r+4)=7

We move all terms to the left:

3(4r+5)-2r(3r+4)-(7)=0

We multiply parentheses

-6r^2+12r-8r+15-7=0

We add all the numbers together, and all the variables

-6r^2+4r+8=0

a = -6; b = 4; c = +8;
Δ = b2-4ac
Δ = 42-4·(-6)·8
Δ = 208
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{13}}{2*-6}=\frac{-4-4\sqrt{13}}{-12}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{13}}{2*-6}=\frac{-4+4\sqrt{13}}{-12}
$