(x+1)(3x+4)=10

Solution for (x+1)(3x+4)=10 equation:

(x+1)(3x+4)=10

We move all terms to the left:

(x+1)(3x+4)-(10)=0

We multiply parentheses ..

(+3x^2+4x+3x+4)-10=0

We get rid of parentheses

3x^2+4x+3x+4-10=0

We add all the numbers together, and all the variables

3x^2+7x-6=0

a = 3; b = 7; c = -6;
Δ = b2-4ac
Δ = 72-4·3·(-6)
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{-(7)-11}{2*3}=\frac{-18}{6}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+11}{2*3}=\frac{4}{6}
=2/3
$