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(4+2x)(7+2x)=38
We move all terms to the left:
(4+2x)(7+2x)-(38)=0
We add all the numbers together, and all the variables
(2x+4)(2x+7)-38=0
We multiply parentheses ..
(+4x^2+14x+8x+28)-38=0
We get rid of parentheses
4x^2+14x+8x+28-38=0
We add all the numbers together, and all the variables
4x^2+22x-10=0
a = 4; b = 22; c = -10;
Δ = b2-4ac
Δ = 222-4·4·(-10)
Δ = 644
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{644}=\sqrt{4*161}=\sqrt{4}*\sqrt{161}=2\sqrt{161}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{161}}{2*4}=\frac{-22-2\sqrt{161}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{161}}{2*4}=\frac{-22+2\sqrt{161}}{8} $
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