3x(x-1)-8=4(1+x)+5

Solution for 3x(x-1)-8=4(1+x)+5 equation:

3x(x-1)-8=4(1+x)+5

We move all terms to the left:

3x(x-1)-8-(4(1+x)+5)=0

We add all the numbers together, and all the variables

3x(x-1)-(4(x+1)+5)-8=0

We multiply parentheses

3x^2-3x-(4(x+1)+5)-8=0


We calculate terms in parentheses: -(4(x+1)+5), so:

4(x+1)+5

We multiply parentheses

4x+4+5

We add all the numbers together, and all the variables

4x+9

Back to the equation:

-(4x+9)


We get rid of parentheses

3x^2-3x-4x-9-8=0

We add all the numbers together, and all the variables

3x^2-7x-17=0

a = 3; b = -7; c = -17;
Δ = b2-4ac
Δ = -72-4·3·(-17)
Δ = 253
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{253}}{2*3}=\frac{7-\sqrt{253}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{253}}{2*3}=\frac{7+\sqrt{253}}{6}
$