0=(h-7)(2h+3)

Solution for 0=(h-7)(2h+3) equation:

0=(h-7)(2h+3)

We move all terms to the left:

0-((h-7)(2h+3))=0

We add all the numbers together, and all the variables

-((h-7)(2h+3))=0

We multiply parentheses ..

-((+2h^2+3h-14h-21))=0


We calculate terms in parentheses: -((+2h^2+3h-14h-21)), so:

(+2h^2+3h-14h-21)

We get rid of parentheses

2h^2+3h-14h-21

We add all the numbers together, and all the variables

2h^2-11h-21

Back to the equation:

-(2h^2-11h-21)


We get rid of parentheses

-2h^2+11h+21=0

a = -2; b = 11; c = +21;
Δ = b2-4ac
Δ = 112-4·(-2)·21
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-17}{2*-2}=\frac{-28}{-4}
=+7
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+17}{2*-2}=\frac{6}{-4}
=-1+1/2
$