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4x^2+44=100
We move all terms to the left:
4x^2+44-(100)=0
We add all the numbers together, and all the variables
4x^2-56=0
a = 4; b = 0; c = -56;
Δ = b2-4ac
Δ = 02-4·4·(-56)
Δ = 896
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{896}=\sqrt{64*14}=\sqrt{64}*\sqrt{14}=8\sqrt{14}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{14}}{2*4}=\frac{0-8\sqrt{14}}{8} =-\frac{8\sqrt{14}}{8} =-\sqrt{14} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{14}}{2*4}=\frac{0+8\sqrt{14}}{8} =\frac{8\sqrt{14}}{8} =\sqrt{14} $
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