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12=5x^2
We move all terms to the left:
12-(5x^2)=0
a = -5; b = 0; c = +12;
Δ = b2-4ac
Δ = 02-4·(-5)·12
Δ = 240
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{15}}{2*-5}=\frac{0-4\sqrt{15}}{-10} =-\frac{4\sqrt{15}}{-10} =-\frac{2\sqrt{15}}{-5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{15}}{2*-5}=\frac{0+4\sqrt{15}}{-10} =\frac{4\sqrt{15}}{-10} =\frac{2\sqrt{15}}{-5} $
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