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16-5x^2+4=10
We move all terms to the left:
16-5x^2+4-(10)=0
We add all the numbers together, and all the variables
-5x^2+10=0
a = -5; b = 0; c = +10;
Δ = b2-4ac
Δ = 02-4·(-5)·10
Δ = 200
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{2}}{2*-5}=\frac{0-10\sqrt{2}}{-10} =-\frac{10\sqrt{2}}{-10} =-\frac{\sqrt{2}}{-1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{2}}{2*-5}=\frac{0+10\sqrt{2}}{-10} =\frac{10\sqrt{2}}{-10} =\frac{\sqrt{2}}{-1} $
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