0=2(x^2+7x-98)

Solution for 0=2(x^2+7x-98) equation:

0=2(x^2+7x-98)

We move all terms to the left:

0-(2(x^2+7x-98))=0

We add all the numbers together, and all the variables

-(2(x^2+7x-98))=0


We calculate terms in parentheses: -(2(x^2+7x-98)), so:

2(x^2+7x-98)

We multiply parentheses

2x^2+14x-196

Back to the equation:

-(2x^2+14x-196)


We get rid of parentheses

-2x^2-14x+196=0

a = -2; b = -14; c = +196;
Δ = b2-4ac
Δ = -142-4·(-2)·196
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-42}{2*-2}=\frac{-28}{-4}
=+7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+42}{2*-2}=\frac{56}{-4}
=-14
$