24^2+h^2=(3h+4)(3h+4)

Solution for 24^2+h^2=(3h+4)(3h+4) equation:

24^2+h^2=(3h+4)(3h+4)

We move all terms to the left:

24^2+h^2-((3h+4)(3h+4))=0

We add all the numbers together, and all the variables

h^2-((3h+4)(3h+4))+576=0

We multiply parentheses ..

h^2-((+9h^2+12h+12h+16))+576=0


We calculate terms in parentheses: -((+9h^2+12h+12h+16)), so:

(+9h^2+12h+12h+16)

We get rid of parentheses

9h^2+12h+12h+16

We add all the numbers together, and all the variables

9h^2+24h+16

Back to the equation:

-(9h^2+24h+16)


We get rid of parentheses

h^2-9h^2-24h-16+576=0

We add all the numbers together, and all the variables

-8h^2-24h+560=0

a = -8; b = -24; c = +560;
Δ = b2-4ac
Δ = -242-4·(-8)·560
Δ = 18496
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{18496}=136$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-136}{2*-8}=\frac{-112}{-16}
=+7
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+136}{2*-8}=\frac{160}{-16}
=-10
$