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b^2/4+45=61
We move all terms to the left:
b^2/4+45-(61)=0
We add all the numbers together, and all the variables
b^2/4-16=0
We multiply all the terms by the denominator
b^2-16*4=0
We add all the numbers together, and all the variables
b^2-64=0
a = 1; b = 0; c = -64;
Δ = b2-4ac
Δ = 02-4·1·(-64)
Δ = 256
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{256}=16$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*1}=\frac{-16}{2} =-8 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*1}=\frac{16}{2} =8 $
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