2x^2+(-3x)=14

Solution for 2x^2+(-3x)=14 equation:

2x^2+(-3x)=14

We move all terms to the left:

2x^2+(-3x)-(14)=0

We get rid of parentheses

2x^2-3x-14=0

a = 2; b = -3; c = -14;
Δ = b2-4ac
Δ = -32-4·2·(-14)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-11}{2*2}=\frac{-8}{4}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+11}{2*2}=\frac{14}{4}
=3+1/2
$