(x+1)(2x+3)=2(x+1)2+8

Solution for (x+1)(2x+3)=2(x+1)2+8 equation:

(x+1)(2x+3)=2(x+1)2+8

We move all terms to the left:

(x+1)(2x+3)-(2(x+1)2+8)=0

We multiply parentheses ..

(+2x^2+3x+2x+3)-(2(x+1)2+8)=0


We calculate terms in parentheses: -(2(x+1)2+8), so:

2(x+1)2+8

We multiply parentheses

4x+4+8

We add all the numbers together, and all the variables

4x+12

Back to the equation:

-(4x+12)


We get rid of parentheses

2x^2+3x+2x-4x+3-12=0

We add all the numbers together, and all the variables

2x^2+x-9=0

a = 2; b = 1; c = -9;
Δ = b2-4ac
Δ = 12-4·2·(-9)
Δ = 73
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{-(1)-\sqrt{73}}{2*2}=\frac{-1-\sqrt{73}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{73}}{2*2}=\frac{-1+\sqrt{73}}{4}
$