98=(2x+4)x

Solution for 98=(2x+4)x equation:

98=(2x+4)x

We move all terms to the left:

98-((2x+4)x)=0


We calculate terms in parentheses: -((2x+4)x), so:

(2x+4)x

We multiply parentheses

2x^2+4x

Back to the equation:

-(2x^2+4x)


We get rid of parentheses

-2x^2-4x+98=0

a = -2; b = -4; c = +98;
Δ = b2-4ac
Δ = -42-4·(-2)·98
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-20\sqrt{2}}{2*-2}=\frac{4-20\sqrt{2}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+20\sqrt{2}}{2*-2}=\frac{4+20\sqrt{2}}{-4}
$