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28x^2+28=32
We move all terms to the left:
28x^2+28-(32)=0
We add all the numbers together, and all the variables
28x^2-4=0
a = 28; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·28·(-4)
Δ = 448
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{7}}{2*28}=\frac{0-8\sqrt{7}}{56} =-\frac{8\sqrt{7}}{56} =-\frac{\sqrt{7}}{7} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{7}}{2*28}=\frac{0+8\sqrt{7}}{56} =\frac{8\sqrt{7}}{56} =\frac{\sqrt{7}}{7} $
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