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a2-10a=23
We move all terms to the left:
a2-10a-(23)=0
We add all the numbers together, and all the variables
a^2-10a-23=0
a = 1; b = -10; c = -23;
Δ = b2-4ac
Δ = -102-4·1·(-23)
Δ = 192
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-8\sqrt{3}}{2*1}=\frac{10-8\sqrt{3}}{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+8\sqrt{3}}{2*1}=\frac{10+8\sqrt{3}}{2} $
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