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(3+2x)(5+2x)=40
We move all terms to the left:
(3+2x)(5+2x)-(40)=0
We add all the numbers together, and all the variables
(2x+3)(2x+5)-40=0
We multiply parentheses ..
(+4x^2+10x+6x+15)-40=0
We get rid of parentheses
4x^2+10x+6x+15-40=0
We add all the numbers together, and all the variables
4x^2+16x-25=0
a = 4; b = 16; c = -25;
Δ = b2-4ac
Δ = 162-4·4·(-25)
Δ = 656
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{41}}{2*4}=\frac{-16-4\sqrt{41}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{41}}{2*4}=\frac{-16+4\sqrt{41}}{8} $
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