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+19x^2=100
We move all terms to the left:
+19x^2-(100)=0
a = 19; b = 0; c = -100;
Δ = b2-4ac
Δ = 02-4·19·(-100)
Δ = 7600
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{7600}=\sqrt{400*19}=\sqrt{400}*\sqrt{19}=20\sqrt{19}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{19}}{2*19}=\frac{0-20\sqrt{19}}{38} =-\frac{20\sqrt{19}}{38} =-\frac{10\sqrt{19}}{19} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{19}}{2*19}=\frac{0+20\sqrt{19}}{38} =\frac{20\sqrt{19}}{38} =\frac{10\sqrt{19}}{19} $
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