8-(h-2)(h+7)=(2+h)(4+h)

Solution for 8-(h-2)(h+7)=(2+h)(4+h) equation:

8-(h-2)(h+7)=(2+h)(4+h)

We move all terms to the left:

8-(h-2)(h+7)-((2+h)(4+h))=0

We add all the numbers together, and all the variables

-(h-2)(h+7)-((h+2)(h+4))+8=0

We multiply parentheses ..

-(+h^2+7h-2h-14)-((h+2)(h+4))+8=0


We calculate terms in parentheses: -((h+2)(h+4)), so:

(h+2)(h+4)

We multiply parentheses ..

(+h^2+4h+2h+8)

We get rid of parentheses

h^2+4h+2h+8

We add all the numbers together, and all the variables

h^2+6h+8

Back to the equation:

-(h^2+6h+8)


We get rid of parentheses

-h^2-h^2-7h+2h-6h+14-8+8=0

We add all the numbers together, and all the variables

-2h^2-11h+14=0

a = -2; b = -11; c = +14;
Δ = b2-4ac
Δ = -112-4·(-2)·14
Δ = 233
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{233}}{2*-2}=\frac{11-\sqrt{233}}{-4}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{233}}{2*-2}=\frac{11+\sqrt{233}}{-4}
$