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16x^2+2=7
We move all terms to the left:
16x^2+2-(7)=0
We add all the numbers together, and all the variables
16x^2-5=0
a = 16; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·16·(-5)
Δ = 320
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{5}}{2*16}=\frac{0-8\sqrt{5}}{32} =-\frac{8\sqrt{5}}{32} =-\frac{\sqrt{5}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{5}}{2*16}=\frac{0+8\sqrt{5}}{32} =\frac{8\sqrt{5}}{32} =\frac{\sqrt{5}}{4} $
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