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10(2)+b(2)=15(2)
We move all terms to the left:
10(2)+b(2)-(15(2))=0
determiningTheFunctionDomain b2+102-152=0
We add all the numbers together, and all the variables
b^2-50=0
a = 1; b = 0; c = -50;
Δ = b2-4ac
Δ = 02-4·1·(-50)
Δ = 200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{2}}{2*1}=\frac{0-10\sqrt{2}}{2} =-\frac{10\sqrt{2}}{2} =-5\sqrt{2} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{2}}{2*1}=\frac{0+10\sqrt{2}}{2} =\frac{10\sqrt{2}}{2} =5\sqrt{2} $
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