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10=m2-3
We move all terms to the left:
10-(m2-3)=0
We add all the numbers together, and all the variables
-(+m^2-3)+10=0
We get rid of parentheses
-m^2+3+10=0
We add all the numbers together, and all the variables
-1m^2+13=0
a = -1; b = 0; c = +13;
Δ = b2-4ac
Δ = 02-4·(-1)·13
Δ = 52
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{13}}{2*-1}=\frac{0-2\sqrt{13}}{-2} =-\frac{2\sqrt{13}}{-2} =-\frac{\sqrt{13}}{-1} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{13}}{2*-1}=\frac{0+2\sqrt{13}}{-2} =\frac{2\sqrt{13}}{-2} =\frac{\sqrt{13}}{-1} $
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