(6a-5)(a+1)=10

Solution for (6a-5)(a+1)=10 equation:

(6a-5)(a+1)=10

We move all terms to the left:

(6a-5)(a+1)-(10)=0

We multiply parentheses ..

(+6a^2+6a-5a-5)-10=0

We get rid of parentheses

6a^2+6a-5a-5-10=0

We add all the numbers together, and all the variables

6a^2+a-15=0

a = 6; b = 1; c = -15;
Δ = b2-4ac
Δ = 12-4·6·(-15)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-19}{2*6}=\frac{-20}{12}
=-1+2/3
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+19}{2*6}=\frac{18}{12}
=1+1/2
$