3x+4x=x^2-70+4x

Solution for 3x+4x=x^2-70+4x equation:

3x+4x=x^2-70+4x

We move all terms to the left:

3x+4x-(x^2-70+4x)=0

We add all the numbers together, and all the variables

7x-(x^2-70+4x)=0

We get rid of parentheses

-x^2+7x-4x+70=0

We add all the numbers together, and all the variables

-1x^2+3x+70=0

a = -1; b = 3; c = +70;
Δ = b2-4ac
Δ = 32-4·(-1)·70
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-17}{2*-1}=\frac{-20}{-2}
=+10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+17}{2*-1}=\frac{14}{-2}
=-7
$