4x+2x^2=(x+35)-3(x^2-x)

Solution for 4x+2x^2=(x+35)-3(x^2-x) equation:

4x+2x^2=(x+35)-3(x^2-x)

We move all terms to the left:

4x+2x^2-((x+35)-3(x^2-x))=0


We calculate terms in parentheses: -((x+35)-3(x^2-x)), so:

(x+35)-3(x^2-x)

We multiply parentheses

-3x^2+(x+35)+3x

We get rid of parentheses

-3x^2+x+3x+35

We add all the numbers together, and all the variables

-3x^2+4x+35

Back to the equation:

-(-3x^2+4x+35)


We get rid of parentheses

2x^2+3x^2-4x+4x-35=0

We add all the numbers together, and all the variables

5x^2-35=0

a = 5; b = 0; c = -35;
Δ = b2-4ac
Δ = 02-4·5·(-35)
Δ = 700
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:

$\sqrt{\Delta}=\sqrt{700}=\sqrt{100*7}=\sqrt{100}*\sqrt{7}=10\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{7}}{2*5}=\frac{0-10\sqrt{7}}{10}
=-\frac{10\sqrt{7}}{10}
=-\sqrt{7}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{7}}{2*5}=\frac{0+10\sqrt{7}}{10}
=\frac{10\sqrt{7}}{10}
=\sqrt{7}
$