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5+2x(x-4)=4x+7
We move all terms to the left:
5+2x(x-4)-(4x+7)=0
We multiply parentheses
2x^2-8x-(4x+7)+5=0
We get rid of parentheses
2x^2-8x-4x-7+5=0
We add all the numbers together, and all the variables
2x^2-12x-2=0
a = 2; b = -12; c = -2;
Δ = b2-4ac
Δ = -122-4·2·(-2)
Δ = 160
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{10}}{2*2}=\frac{12-4\sqrt{10}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{10}}{2*2}=\frac{12+4\sqrt{10}}{4} $
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