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m^2-7=88
We move all terms to the left:
m^2-7-(88)=0
We add all the numbers together, and all the variables
m^2-95=0
a = 1; b = 0; c = -95;
Δ = b2-4ac
Δ = 02-4·1·(-95)
Δ = 380
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{380}=\sqrt{4*95}=\sqrt{4}*\sqrt{95}=2\sqrt{95}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{95}}{2*1}=\frac{0-2\sqrt{95}}{2} =-\frac{2\sqrt{95}}{2} =-\sqrt{95} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{95}}{2*1}=\frac{0+2\sqrt{95}}{2} =\frac{2\sqrt{95}}{2} =\sqrt{95} $
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