2X^2-19x^2+32X+21=03

Solution for 2X^2-19x^2+32X+21=03 equation:

2X^2-19X^2+32X+21=03

We move all terms to the left:

2X^2-19X^2+32X+21-(03)=0

We add all the numbers together, and all the variables

-17X^2+32X+18=0

a = -17; b = 32; c = +18;
Δ = b2-4ac
Δ = 322-4·(-17)·18
Δ = 2248
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:

$\sqrt{\Delta}=\sqrt{2248}=\sqrt{4*562}=\sqrt{4}*\sqrt{562}=2\sqrt{562}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-2\sqrt{562}}{2*-17}=\frac{-32-2\sqrt{562}}{-34}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+2\sqrt{562}}{2*-17}=\frac{-32+2\sqrt{562}}{-34}
$