H(x)=(2x+1)(x-3)

Solution for H(x)=(2x+1)(x-3) equation:

(H)=(2H+1)(H-3)

We move all terms to the left:

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

We multiply parentheses ..

-((+2H^2-6H+H-3))+H=0


We calculate terms in parentheses: -((+2H^2-6H+H-3)), so:

(+2H^2-6H+H-3)

We get rid of parentheses

2H^2-6H+H-3

We add all the numbers together, and all the variables

2H^2-5H-3

Back to the equation:

-(2H^2-5H-3)


We add all the numbers together, and all the variables

H-(2H^2-5H-3)=0

We get rid of parentheses

-2H^2+H+5H+3=0

We add all the numbers together, and all the variables

-2H^2+6H+3=0

a = -2; b = 6; c = +3;
Δ = b2-4ac
Δ = 62-4·(-2)·3
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{15}}{2*-2}=\frac{-6-2\sqrt{15}}{-4}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{15}}{2*-2}=\frac{-6+2\sqrt{15}}{-4}
$