A(4)=(2x+10)(2x+15)

Solution for A(4)=(2x+10)(2x+15) equation:

(4)=(2A+10)(2A+15)

We move all terms to the left:

(4)-((2A+10)(2A+15))=0

We multiply parentheses ..

-((+4A^2+30A+20A+150))+4=0


We calculate terms in parentheses: -((+4A^2+30A+20A+150)), so:

(+4A^2+30A+20A+150)

We get rid of parentheses

4A^2+30A+20A+150

We add all the numbers together, and all the variables

4A^2+50A+150

Back to the equation:

-(4A^2+50A+150)


We get rid of parentheses

-4A^2-50A-150+4=0

We add all the numbers together, and all the variables

-4A^2-50A-146=0

a = -4; b = -50; c = -146;
Δ = b2-4ac
Δ = -502-4·(-4)·(-146)
Δ = 164
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-2\sqrt{41}}{2*-4}=\frac{50-2\sqrt{41}}{-8}
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+2\sqrt{41}}{2*-4}=\frac{50+2\sqrt{41}}{-8}
$