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b^2-24=-5b
We move all terms to the left:
b^2-24-(-5b)=0
We get rid of parentheses
b^2+5b-24=0
a = 1; b = 5; c = -24;
Δ = b2-4ac
Δ = 52-4·1·(-24)
Δ = 121
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{121}=11$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-11}{2*1}=\frac{-16}{2} =-8 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+11}{2*1}=\frac{6}{2} =3 $
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