4x^2-7x-8=2x(x+1)-3

Solution for 4x^2-7x-8=2x(x+1)-3 equation:

4x^2-7x-8=2x(x+1)-3

We move all terms to the left:

4x^2-7x-8-(2x(x+1)-3)=0


We calculate terms in parentheses: -(2x(x+1)-3), so:

2x(x+1)-3

We multiply parentheses

2x^2+2x-3

Back to the equation:

-(2x^2+2x-3)


We get rid of parentheses

4x^2-2x^2-7x-2x+3-8=0

We add all the numbers together, and all the variables

2x^2-9x-5=0

a = 2; b = -9; c = -5;
Δ = b2-4ac
Δ = -92-4·2·(-5)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-11}{2*2}=\frac{-2}{4}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+11}{2*2}=\frac{20}{4}
=5
$