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6-5a^2=0
a = -5; b = 0; c = +6;
Δ = b2-4ac
Δ = 02-4·(-5)·6
Δ = 120
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{30}}{2*-5}=\frac{0-2\sqrt{30}}{-10} =-\frac{2\sqrt{30}}{-10} =-\frac{\sqrt{30}}{-5} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{30}}{2*-5}=\frac{0+2\sqrt{30}}{-10} =\frac{2\sqrt{30}}{-10} =\frac{\sqrt{30}}{-5} $
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