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x^2-3x-25=4
We move all terms to the left:
x^2-3x-25-(4)=0
We add all the numbers together, and all the variables
x^2-3x-29=0
a = 1; b = -3; c = -29;
Δ = b2-4ac
Δ = -32-4·1·(-29)
Δ = 125
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{125}=\sqrt{25*5}=\sqrt{25}*\sqrt{5}=5\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5\sqrt{5}}{2*1}=\frac{3-5\sqrt{5}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5\sqrt{5}}{2*1}=\frac{3+5\sqrt{5}}{2} $
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