5(r)=2r^2-4

Solution for 5(r)=2r^2-4 equation:

5(r)=2r^2-4

We move all terms to the left:

5(r)-(2r^2-4)=0

We get rid of parentheses

-2r^2+5r+4=0

a = -2; b = 5; c = +4;
Δ = b2-4ac
Δ = 52-4·(-2)·4
Δ = 57
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{57}}{2*-2}=\frac{-5-\sqrt{57}}{-4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{57}}{2*-2}=\frac{-5+\sqrt{57}}{-4}
$