A=(6x+10)(8x-3)

Solution for A=(6x+10)(8x-3) equation:

=(6A+10)(8A-3)

We move all terms to the left:

-((6A+10)(8A-3))=0

We multiply parentheses ..

-((+48A^2-18A+80A-30))=0


We calculate terms in parentheses: -((+48A^2-18A+80A-30)), so:

(+48A^2-18A+80A-30)

We get rid of parentheses

48A^2-18A+80A-30

We add all the numbers together, and all the variables

48A^2+62A-30

Back to the equation:

-(48A^2+62A-30)


We get rid of parentheses

-48A^2-62A+30=0

a = -48; b = -62; c = +30;
Δ = b2-4ac
Δ = -622-4·(-48)·30
Δ = 9604
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9604}=98$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-62)-98}{2*-48}=\frac{-36}{-96}
=3/8
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-62)+98}{2*-48}=\frac{160}{-96}
=-1+2/3
$