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5m^2-23-42=0
We add all the numbers together, and all the variables
5m^2-65=0
a = 5; b = 0; c = -65;
Δ = b2-4ac
Δ = 02-4·5·(-65)
Δ = 1300
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1300}=\sqrt{100*13}=\sqrt{100}*\sqrt{13}=10\sqrt{13}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{13}}{2*5}=\frac{0-10\sqrt{13}}{10} =-\frac{10\sqrt{13}}{10} =-\sqrt{13} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{13}}{2*5}=\frac{0+10\sqrt{13}}{10} =\frac{10\sqrt{13}}{10} =\sqrt{13} $
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