H(x)=-4(x-3)(x-1)

Solution for H(x)=-4(x-3)(x-1) equation:

(H)=-4(H-3)(H-1)

We move all terms to the left:

(H)-(-4(H-3)(H-1))=0

We multiply parentheses ..

-(-4(+H^2-1H-3H+3))+H=0


We calculate terms in parentheses: -(-4(+H^2-1H-3H+3)), so:

-4(+H^2-1H-3H+3)

We multiply parentheses

-4H^2+4H+12H-12

We add all the numbers together, and all the variables

-4H^2+16H-12

Back to the equation:

-(-4H^2+16H-12)


We get rid of parentheses

4H^2-16H+H+12=0

We add all the numbers together, and all the variables

4H^2-15H+12=0

a = 4; b = -15; c = +12;
Δ = b2-4ac
Δ = -152-4·4·12
Δ = 33
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{-(-15)-\sqrt{33}}{2*4}=\frac{15-\sqrt{33}}{8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{33}}{2*4}=\frac{15+\sqrt{33}}{8}
$