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

Solution for 37=x^2+(3x-5)+(x+2)= equation:

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

We move all terms to the left:

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


We calculate terms in parentheses: -(x^2+(3x-5)+(x+2)), so:

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

We get rid of parentheses

x^2+3x+x-5+2

We add all the numbers together, and all the variables

x^2+4x-3

Back to the equation:

-(x^2+4x-3)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-1x^2-4x+40=0

a = -1; b = -4; c = +40;
Δ = b2-4ac
Δ = -42-4·(-1)·40
Δ = 176
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:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{11}}{2*-1}=\frac{4-4\sqrt{11}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{11}}{2*-1}=\frac{4+4\sqrt{11}}{-2}
$