3(2h+9)=2h(4h-5)

Solution for 3(2h+9)=2h(4h-5) equation:

3(2h+9)=2h(4h-5)

We move all terms to the left:

3(2h+9)-(2h(4h-5))=0

We multiply parentheses

6h-(2h(4h-5))+27=0


We calculate terms in parentheses: -(2h(4h-5)), so:

2h(4h-5)

We multiply parentheses

8h^2-10h

Back to the equation:

-(8h^2-10h)


We get rid of parentheses

-8h^2+6h+10h+27=0

We add all the numbers together, and all the variables

-8h^2+16h+27=0

a = -8; b = 16; c = +27;
Δ = b2-4ac
Δ = 162-4·(-8)·27
Δ = 1120
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1120}=\sqrt{16*70}=\sqrt{16}*\sqrt{70}=4\sqrt{70}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{70}}{2*-8}=\frac{-16-4\sqrt{70}}{-16}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{70}}{2*-8}=\frac{-16+4\sqrt{70}}{-16}
$