3x2+7x-2=(x+1)2

Solution for 3x2+7x-2=(x+1)2 equation:

3x^2+7x-2=(x+1)2

We move all terms to the left:

3x^2+7x-2-((x+1)2)=0


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

(x+1)2

We multiply parentheses

2x+2

Back to the equation:

-(2x+2)


We get rid of parentheses

3x^2+7x-2x-2-2=0

We add all the numbers together, and all the variables

3x^2+5x-4=0

a = 3; b = 5; c = -4;
Δ = b2-4ac
Δ = 52-4·3·(-4)
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{-(5)-\sqrt{73}}{2*3}=\frac{-5-\sqrt{73}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{73}}{2*3}=\frac{-5+\sqrt{73}}{6}
$