2x^2=7(x+1)

Solution for 2x^2=7(x+1) equation:

2x^2=7(x+1)

We move all terms to the left:

2x^2-(7(x+1))=0


We calculate terms in parentheses: -(7(x+1)), so:

7(x+1)

We multiply parentheses

7x+7

Back to the equation:

-(7x+7)


We get rid of parentheses

2x^2-7x-7=0

a = 2; b = -7; c = -7;
Δ = b2-4ac
Δ = -72-4·2·(-7)
Δ = 105
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{-(-7)-\sqrt{105}}{2*2}=\frac{7-\sqrt{105}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{105}}{2*2}=\frac{7+\sqrt{105}}{4}
$