(2x-2)(x+12)=(x+98)

Solution for (2x-2)(x+12)=(x+98) equation:

(2x-2)(x+12)=(x+98)

We move all terms to the left:

(2x-2)(x+12)-((x+98))=0

We multiply parentheses ..

(+2x^2+24x-2x-24)-((x+98))=0


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

(x+98)

We get rid of parentheses

x+98

Back to the equation:

-(x+98)


We get rid of parentheses

2x^2+24x-2x-x-24-98=0

We add all the numbers together, and all the variables

2x^2+21x-122=0

a = 2; b = 21; c = -122;
Δ = b2-4ac
Δ = 212-4·2·(-122)
Δ = 1417
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-\sqrt{1417}}{2*2}=\frac{-21-\sqrt{1417}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{1417}}{2*2}=\frac{-21+\sqrt{1417}}{4}
$