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

Solution for (2x^2+x-4)+(-3^2-5x=1) equation:

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-4x-14=0

a = 2; b = -4; c = -14;
Δ = b2-4ac
Δ = -42-4·2·(-14)
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8\sqrt{2}}{2*2}=\frac{4-8\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8\sqrt{2}}{2*2}=\frac{4+8\sqrt{2}}{4}
$