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4x^2+42x=2x-8
We move all terms to the left:
4x^2+42x-(2x-8)=0
We get rid of parentheses
4x^2+42x-2x+8=0
We add all the numbers together, and all the variables
4x^2+40x+8=0
a = 4; b = 40; c = +8;
Δ = b2-4ac
Δ = 402-4·4·8
Δ = 1472
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1472}=\sqrt{64*23}=\sqrt{64}*\sqrt{23}=8\sqrt{23}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{23}}{2*4}=\frac{-40-8\sqrt{23}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{23}}{2*4}=\frac{-40+8\sqrt{23}}{8} $
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