-9r^2-23=-10r^2-10+3r

Solution for -9r^2-23=-10r^2-10+3r equation:

-9r^2-23=-10r^2-10+3r

We move all terms to the left:

-9r^2-23-(-10r^2-10+3r)=0

We get rid of parentheses

-9r^2+10r^2-3r+10-23=0

We add all the numbers together, and all the variables

r^2-3r-13=0

a = 1; b = -3; c = -13;
Δ = b2-4ac
Δ = -32-4·1·(-13)
Δ = 61
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{-(-3)-\sqrt{61}}{2*1}=\frac{3-\sqrt{61}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{61}}{2*1}=\frac{3+\sqrt{61}}{2}
$