(x-4)(x-3)+x=37

Solution for (x-4)(x-3)+x=37 equation:

(x-4)(x-3)+x=37

We move all terms to the left:

(x-4)(x-3)+x-(37)=0

We add all the numbers together, and all the variables

x+(x-4)(x-3)-37=0

We multiply parentheses ..

(+x^2-3x-4x+12)+x-37=0

We get rid of parentheses

x^2-3x-4x+x+12-37=0

We add all the numbers together, and all the variables

x^2-6x-25=0

a = 1; b = -6; c = -25;
Δ = b2-4ac
Δ = -62-4·1·(-25)
Δ = 136
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:

$\sqrt{\Delta}=\sqrt{136}=\sqrt{4*34}=\sqrt{4}*\sqrt{34}=2\sqrt{34}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{34}}{2*1}=\frac{6-2\sqrt{34}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{34}}{2*1}=\frac{6+2\sqrt{34}}{2}
$