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m^2-46=0
a = 1; b = 0; c = -46;
Δ = b2-4ac
Δ = 02-4·1·(-46)
Δ = 184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{46}}{2*1}=\frac{0-2\sqrt{46}}{2} =-\frac{2\sqrt{46}}{2} =-\sqrt{46} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{46}}{2*1}=\frac{0+2\sqrt{46}}{2} =\frac{2\sqrt{46}}{2} =\sqrt{46} $
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