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3a^2-12+2a^2-10+12=0
We add all the numbers together, and all the variables
5a^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:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{2}}{2*5}=\frac{0-10\sqrt{2}}{10} =-\frac{10\sqrt{2}}{10} =-\sqrt{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{2}}{2*5}=\frac{0+10\sqrt{2}}{10} =\frac{10\sqrt{2}}{10} =\sqrt{2} $
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