0=2(x^2+5x-14)

Solution for 0=2(x^2+5x-14) equation:

0=2(x^2+5x-14)

We move all terms to the left:

0-(2(x^2+5x-14))=0

We add all the numbers together, and all the variables

-(2(x^2+5x-14))=0


We calculate terms in parentheses: -(2(x^2+5x-14)), so:

2(x^2+5x-14)

We multiply parentheses

2x^2+10x-28

Back to the equation:

-(2x^2+10x-28)


We get rid of parentheses

-2x^2-10x+28=0

a = -2; b = -10; c = +28;
Δ = b2-4ac
Δ = -102-4·(-2)·28
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-18}{2*-2}=\frac{-8}{-4}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+18}{2*-2}=\frac{28}{-4}
=-7
$