10(x+1)=5(2x2)

Solution for 10(x+1)=5(2x2) equation:

10(x+1)=5(2x^2)

We move all terms to the left:

10(x+1)-(5(2x^2))=0

determiningTheFunctionDomain
10(x+1)-52x^2=0

We add all the numbers together, and all the variables

-52x^2+10(x+1)=0

We multiply parentheses

-52x^2+10x+10=0

a = -52; b = 10; c = +10;
Δ = b2-4ac
Δ = 102-4·(-52)·10
Δ = 2180
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:

$\sqrt{\Delta}=\sqrt{2180}=\sqrt{4*545}=\sqrt{4}*\sqrt{545}=2\sqrt{545}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{545}}{2*-52}=\frac{-10-2\sqrt{545}}{-104}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{545}}{2*-52}=\frac{-10+2\sqrt{545}}{-104}
$