x+x2+(x-3)=61

Solution for x+x2+(x-3)=61 equation:

x+x2+(x-3)=61

We move all terms to the left:

x+x2+(x-3)-(61)=0

We add all the numbers together, and all the variables

x^2+x+(x-3)-61=0

We get rid of parentheses

x^2+x+x-3-61=0

We add all the numbers together, and all the variables

x^2+2x-64=0

a = 1; b = 2; c = -64;
Δ = b2-4ac
Δ = 22-4·1·(-64)
Δ = 260
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:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{65}}{2*1}=\frac{-2-2\sqrt{65}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{65}}{2*1}=\frac{-2+2\sqrt{65}}{2}
$