(r-3)(3r+4)=-10

Solution for (r-3)(3r+4)=-10 equation:

(r-3)(3r+4)=-10

We move all terms to the left:

(r-3)(3r+4)-(-10)=0

We add all the numbers together, and all the variables

(r-3)(3r+4)+10=0

We multiply parentheses ..

(+3r^2+4r-9r-12)+10=0

We get rid of parentheses

3r^2+4r-9r-12+10=0

We add all the numbers together, and all the variables

3r^2-5r-2=0

a = 3; b = -5; c = -2;
Δ = b2-4ac
Δ = -52-4·3·(-2)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-7}{2*3}=\frac{-2}{6}
=-1/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+7}{2*3}=\frac{12}{6}
=2
$