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28x^2+28=52
We move all terms to the left:
28x^2+28-(52)=0
We add all the numbers together, and all the variables
28x^2-24=0
a = 28; b = 0; c = -24;
Δ = b2-4ac
Δ = 02-4·28·(-24)
Δ = 2688
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2688}=\sqrt{64*42}=\sqrt{64}*\sqrt{42}=8\sqrt{42}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{42}}{2*28}=\frac{0-8\sqrt{42}}{56} =-\frac{8\sqrt{42}}{56} =-\frac{\sqrt{42}}{7} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{42}}{2*28}=\frac{0+8\sqrt{42}}{56} =\frac{8\sqrt{42}}{56} =\frac{\sqrt{42}}{7} $
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