a2+9=49

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Solution for a2+9=49 equation:



a2+9=49
We move all terms to the left:
a2+9-(49)=0
We add all the numbers together, and all the variables
a^2-40=0
a = 1; b = 0; c = -40;
Δ = b2-4ac
Δ = 02-4·1·(-40)
Δ = 160
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{10}}{2*1}=\frac{0-4\sqrt{10}}{2} =-\frac{4\sqrt{10}}{2} =-2\sqrt{10} $
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{10}}{2*1}=\frac{0+4\sqrt{10}}{2} =\frac{4\sqrt{10}}{2} =2\sqrt{10} $

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