(x^2)+(3x/20)=90

Solution for (x^2)+(3x/20)=90 equation:

(x^2)+(3x/20)=90

We move all terms to the left:

(x^2)+(3x/20)-(90)=0

We add all the numbers together, and all the variables

x^2+(+3x/20)-90=0

We get rid of parentheses

x^2+3x/20-90=0

We multiply all the terms by the denominator


x^2*20+3x-90*20=0

We add all the numbers together, and all the variables

x^2*20+3x-1800=0

Wy multiply elements

20x^2+3x-1800=0

a = 20; b = 3; c = -1800;
Δ = b2-4ac
Δ = 32-4·20·(-1800)
Δ = 144009
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{144009}=\sqrt{9*16001}=\sqrt{9}*\sqrt{16001}=3\sqrt{16001}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{16001}}{2*20}=\frac{-3-3\sqrt{16001}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{16001}}{2*20}=\frac{-3+3\sqrt{16001}}{40}
$