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x^2+10x+21=38
We move all terms to the left:
x^2+10x+21-(38)=0
We add all the numbers together, and all the variables
x^2+10x-17=0
a = 1; b = 10; c = -17;
Δ = b2-4ac
Δ = 102-4·1·(-17)
Δ = 168
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{42}}{2*1}=\frac{-10-2\sqrt{42}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{42}}{2*1}=\frac{-10+2\sqrt{42}}{2} $
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