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15x^2+2x-10=0
a = 15; b = 2; c = -10;
Δ = b2-4ac
Δ = 22-4·15·(-10)
Δ = 604
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{604}=\sqrt{4*151}=\sqrt{4}*\sqrt{151}=2\sqrt{151}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{151}}{2*15}=\frac{-2-2\sqrt{151}}{30} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{151}}{2*15}=\frac{-2+2\sqrt{151}}{30} $
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