2x^+4(x^2+3)-2=22

Solution for 2x^+4(x^2+3)-2=22 equation:

2x^+4(x^2+3)-2=22

We move all terms to the left:

2x^+4(x^2+3)-2-(22)=0

We add all the numbers together, and all the variables

2x+4(x^2+3)-24=0

We multiply parentheses

4x^2+2x+12-24=0

We add all the numbers together, and all the variables

4x^2+2x-12=0

a = 4; b = 2; c = -12;
Δ = b2-4ac
Δ = 22-4·4·(-12)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-14}{2*4}=\frac{-16}{8}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+14}{2*4}=\frac{12}{8}
=1+1/2
$