x^2+2x=4x+98

Solution for x^2+2x=4x+98 equation:

x^2+2x=4x+98

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-2x-98=0

a = 1; b = -2; c = -98;
Δ = b2-4ac
Δ = -22-4·1·(-98)
Δ = 396
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:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6\sqrt{11}}{2*1}=\frac{2-6\sqrt{11}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6\sqrt{11}}{2*1}=\frac{2+6\sqrt{11}}{2}
$