-4r^2+10r+9=-2r^2

Solution for -4r^2+10r+9=-2r^2 equation:

-4r^2+10r+9=-2r^2

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-2r^2+10r+9=0

a = -2; b = 10; c = +9;
Δ = b2-4ac
Δ = 102-4·(-2)·9
Δ = 172
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:

$\sqrt{\Delta}=\sqrt{172}=\sqrt{4*43}=\sqrt{4}*\sqrt{43}=2\sqrt{43}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{43}}{2*-2}=\frac{-10-2\sqrt{43}}{-4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{43}}{2*-2}=\frac{-10+2\sqrt{43}}{-4}
$