x^2-4(3-x)+5x=4

Solution for x^2-4(3-x)+5x=4 equation:

x^2-4(3-x)+5x=4

We move all terms to the left:

x^2-4(3-x)+5x-(4)=0

We add all the numbers together, and all the variables

x^2-4(-1x+3)+5x-4=0

We add all the numbers together, and all the variables

x^2+5x-4(-1x+3)-4=0

We multiply parentheses

x^2+5x+4x-12-4=0

We add all the numbers together, and all the variables

x^2+9x-16=0

a = 1; b = 9; c = -16;
Δ = b2-4ac
Δ = 92-4·1·(-16)
Δ = 145
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{145}}{2*1}=\frac{-9-\sqrt{145}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{145}}{2*1}=\frac{-9+\sqrt{145}}{2}
$