x+2x^2-15=21

Solution for x+2x^2-15=21 equation:

x+2x^2-15=21

We move all terms to the left:

x+2x^2-15-(21)=0

We add all the numbers together, and all the variables

2x^2+x-36=0

a = 2; b = 1; c = -36;
Δ = b2-4ac
Δ = 12-4·2·(-36)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-17}{2*2}=\frac{-18}{4}
=-4+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+17}{2*2}=\frac{16}{4}
=4
$