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b2+5=8b-10
We move all terms to the left:
b2+5-(8b-10)=0
We add all the numbers together, and all the variables
b^2-(8b-10)+5=0
We get rid of parentheses
b^2-8b+10+5=0
We add all the numbers together, and all the variables
b^2-8b+15=0
a = 1; b = -8; c = +15;
Δ = b2-4ac
Δ = -82-4·1·15
Δ = 4
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}\sqrt{\Delta}=\sqrt{4}=2b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2}{2*1}=\frac{6}{2} =3b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2}{2*1}=\frac{10}{2} =5
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