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32+40x^2=352
We move all terms to the left:
32+40x^2-(352)=0
We add all the numbers together, and all the variables
40x^2-320=0
a = 40; b = 0; c = -320;
Δ = b2-4ac
Δ = 02-4·40·(-320)
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{-(0)-160\sqrt{2}}{2*40}=\frac{0-160\sqrt{2}}{80} =-\frac{160\sqrt{2}}{80} =-2\sqrt{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+160\sqrt{2}}{2*40}=\frac{0+160\sqrt{2}}{80} =\frac{160\sqrt{2}}{80} =2\sqrt{2} $
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