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40x^2+240x+200=160
We move all terms to the left:
40x^2+240x+200-(160)=0
We add all the numbers together, and all the variables
40x^2+240x+40=0
a = 40; b = 240; c = +40;
Δ = b2-4ac
Δ = 2402-4·40·40
Δ = 51200
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{51200}=\sqrt{25600*2}=\sqrt{25600}*\sqrt{2}=160\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(240)-160\sqrt{2}}{2*40}=\frac{-240-160\sqrt{2}}{80} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(240)+160\sqrt{2}}{2*40}=\frac{-240+160\sqrt{2}}{80} $
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