(X+5)(3x+4)=3(x+2)-2

Solution for (X+5)(3x+4)=3(x+2)-2 equation:

(X+5)(3X+4)=3(X+2)-2

We move all terms to the left:

(X+5)(3X+4)-(3(X+2)-2)=0

We multiply parentheses ..

(+3X^2+4X+15X+20)-(3(X+2)-2)=0


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

3(X+2)-2

We multiply parentheses

3X+6-2

We add all the numbers together, and all the variables

3X+4

Back to the equation:

-(3X+4)


We get rid of parentheses

3X^2+4X+15X-3X+20-4=0

We add all the numbers together, and all the variables

3X^2+16X+16=0

a = 3; b = 16; c = +16;
Δ = b2-4ac
Δ = 162-4·3·16
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8}{2*3}=\frac{-24}{6}
=-4
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8}{2*3}=\frac{-8}{6}
=-1+1/3
$