0=2(x^2+7x-44)

Solution for 0=2(x^2+7x-44) equation:

0=2(x^2+7x-44)

We move all terms to the left:

0-(2(x^2+7x-44))=0

We add all the numbers together, and all the variables

-(2(x^2+7x-44))=0


We calculate terms in parentheses: -(2(x^2+7x-44)), so:

2(x^2+7x-44)

We multiply parentheses

2x^2+14x-88

Back to the equation:

-(2x^2+14x-88)


We get rid of parentheses

-2x^2-14x+88=0

a = -2; b = -14; c = +88;
Δ = b2-4ac
Δ = -142-4·(-2)·88
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-30}{2*-2}=\frac{-16}{-4}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+30}{2*-2}=\frac{44}{-4}
=-11
$