35=70-34x+4x^2

Solution for 35=70-34x+4x^2 equation:

35=70-34x+4x^2

We move all terms to the left:

35-(70-34x+4x^2)=0

We get rid of parentheses

-4x^2+34x-70+35=0

We add all the numbers together, and all the variables

-4x^2+34x-35=0

a = -4; b = 34; c = -35;
Δ = b2-4ac
Δ = 342-4·(-4)·(-35)
Δ = 596
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:

$\sqrt{\Delta}=\sqrt{596}=\sqrt{4*149}=\sqrt{4}*\sqrt{149}=2\sqrt{149}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-2\sqrt{149}}{2*-4}=\frac{-34-2\sqrt{149}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+2\sqrt{149}}{2*-4}=\frac{-34+2\sqrt{149}}{-8}
$