(3r^2-12r-63)=0

Solution for (3r^2-12r-63)=0 equation:

(3r^2-12r-63)=0

We get rid of parentheses

3r^2-12r-63=0

a = 3; b = -12; c = -63;
Δ = b2-4ac
Δ = -122-4·3·(-63)
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-30}{2*3}=\frac{-18}{6}
=-3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+30}{2*3}=\frac{42}{6}
=7
$