H(x)=3/5x+2

Solution for H(x)=3/5x+2 equation:

(H)=3/5H+2

We move all terms to the left:

(H)-(3/5H+2)=0


Domain of the equation: 5H+2)!=0

H∈R


We get rid of parentheses

H-3/5H-2=0

We multiply all the terms by the denominator


H*5H-2*5H-3=0

Wy multiply elements

5H^2-10H-3=0

a = 5; b = -10; c = -3;
Δ = b2-4ac
Δ = -102-4·5·(-3)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-4\sqrt{10}}{2*5}=\frac{10-4\sqrt{10}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+4\sqrt{10}}{2*5}=\frac{10+4\sqrt{10}}{10}
$