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

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

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

We move all terms to the left:

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

We multiply parentheses

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

We add all the numbers together, and all the variables

12r^2+9r-15=0

a = 12; b = 9; c = -15;
Δ = b2-4ac
Δ = 92-4·12·(-15)
Δ = 801
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:

$\sqrt{\Delta}=\sqrt{801}=\sqrt{9*89}=\sqrt{9}*\sqrt{89}=3\sqrt{89}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{89}}{2*12}=\frac{-9-3\sqrt{89}}{24}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{89}}{2*12}=\frac{-9+3\sqrt{89}}{24}
$