4r+48=4r+24+4r2-8r

Solution for 4r+48=4r+24+4r2-8r equation:

4r+48=4r+24+4r^2-8r

We move all terms to the left:

4r+48-(4r+24+4r^2-8r)=0

We get rid of parentheses

-4r^2-4r+8r+4r-24+48=0

We add all the numbers together, and all the variables

-4r^2+8r+24=0

a = -4; b = 8; c = +24;
Δ = b2-4ac
Δ = 82-4·(-4)·24
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{7}}{2*-4}=\frac{-8-8\sqrt{7}}{-8}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{7}}{2*-4}=\frac{-8+8\sqrt{7}}{-8}
$