H(5x^2-3x-4)=2x+7

Solution for H(5x^2-3x-4)=2x+7 equation:

(5H^2-3H-4)=2H+7

We move all terms to the left:

(5H^2-3H-4)-(2H+7)=0

We get rid of parentheses

5H^2-3H-2H-4-7=0

We add all the numbers together, and all the variables

5H^2-5H-11=0

a = 5; b = -5; c = -11;
Δ = b2-4ac
Δ = -52-4·5·(-11)
Δ = 245
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:

$\sqrt{\Delta}=\sqrt{245}=\sqrt{49*5}=\sqrt{49}*\sqrt{5}=7\sqrt{5}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-7\sqrt{5}}{2*5}=\frac{5-7\sqrt{5}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+7\sqrt{5}}{2*5}=\frac{5+7\sqrt{5}}{10}
$