2x^2-10x=38x-98

Solution for 2x^2-10x=38x-98 equation:

2x^2-10x=38x-98

We move all terms to the left:

2x^2-10x-(38x-98)=0

We get rid of parentheses

2x^2-10x-38x+98=0

We add all the numbers together, and all the variables

2x^2-48x+98=0

a = 2; b = -48; c = +98;
Δ = b2-4ac
Δ = -482-4·2·98
Δ = 1520
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:

$\sqrt{\Delta}=\sqrt{1520}=\sqrt{16*95}=\sqrt{16}*\sqrt{95}=4\sqrt{95}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-4\sqrt{95}}{2*2}=\frac{48-4\sqrt{95}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+4\sqrt{95}}{2*2}=\frac{48+4\sqrt{95}}{4}
$