(10x^2/2+5x)-20=0

Solution for (10x^2/2+5x)-20=0 equation:

(10x^2/2+5x)-20=0


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

We move all terms containing x to the left, all other terms to the right

5x)!=-2

x!=-2/1

x!=-2

x∈R


We get rid of parentheses

10x^2/2+5x-20=0

We multiply all the terms by the denominator


10x^2+5x*2-20*2=0

We add all the numbers together, and all the variables

10x^2+5x*2-40=0

Wy multiply elements

10x^2+10x-40=0

a = 10; b = 10; c = -40;
Δ = b2-4ac
Δ = 102-4·10·(-40)
Δ = 1700
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:

$\sqrt{\Delta}=\sqrt{1700}=\sqrt{100*17}=\sqrt{100}*\sqrt{17}=10\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{17}}{2*10}=\frac{-10-10\sqrt{17}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{17}}{2*10}=\frac{-10+10\sqrt{17}}{20}
$