(3h+h)(4h-7)=0

Solution for (3h+h)(4h-7)=0 equation:

(3h+h)(4h-7)=0

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+16h^2-28h)=0

We get rid of parentheses

16h^2-28h=0

a = 16; b = -28; c = 0;
Δ = b2-4ac
Δ = -282-4·16·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-28}{2*16}=\frac{0}{32}
=0
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+28}{2*16}=\frac{56}{32}
=1+3/4
$