Q(r)=3r^2-15r+18

Solution for Q(r)=3r^2-15r+18 equation:

(Q)=3Q^2-15Q+18

We move all terms to the left:

(Q)-(3Q^2-15Q+18)=0

We get rid of parentheses

-3Q^2+Q+15Q-18=0

We add all the numbers together, and all the variables

-3Q^2+16Q-18=0

a = -3; b = 16; c = -18;
Δ = b2-4ac
Δ = 162-4·(-3)·(-18)
Δ = 40
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$Q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{10}}{2*-3}=\frac{-16-2\sqrt{10}}{-6}
$
$Q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{10}}{2*-3}=\frac{-16+2\sqrt{10}}{-6}
$