3x+(4x-5);x=10

Solution for 3x+(4x-5);x=10 equation:

3x+(4x-5)x=10

We move all terms to the left:

3x+(4x-5)x-(10)=0

We multiply parentheses

4x^2+3x-5x-10=0

We add all the numbers together, and all the variables

4x^2-2x-10=0

a = 4; b = -2; c = -10;
Δ = b2-4ac
Δ = -22-4·4·(-10)
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{41}}{2*4}=\frac{2-2\sqrt{41}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{41}}{2*4}=\frac{2+2\sqrt{41}}{8}
$