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X^2-4X-5=300
We move all terms to the left:
X^2-4X-5-(300)=0
We add all the numbers together, and all the variables
X^2-4X-305=0
a = 1; b = -4; c = -305;
Δ = b2-4ac
Δ = -42-4·1·(-305)
Δ = 1236
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1236}=\sqrt{4*309}=\sqrt{4}*\sqrt{309}=2\sqrt{309}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{309}}{2*1}=\frac{4-2\sqrt{309}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{309}}{2*1}=\frac{4+2\sqrt{309}}{2} $
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