(4h+1)(h-7)=0

Solution for (4h+1)(h-7)=0 equation:

(4h+1)(h-7)=0

We multiply parentheses ..

(+4h^2-28h+h-7)=0

We get rid of parentheses

4h^2-28h+h-7=0

We add all the numbers together, and all the variables

4h^2-27h-7=0

a = 4; b = -27; c = -7;
Δ = b2-4ac
Δ = -272-4·4·(-7)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-29}{2*4}=\frac{-2}{8}
=-1/4
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+29}{2*4}=\frac{56}{8}
=7
$