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15x^2+42x+4=00
a = 15; b = 42; c = +4;
Δ = b2-4ac
Δ = 422-4·15·4
Δ = 1524
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1524}=\sqrt{4*381}=\sqrt{4}*\sqrt{381}=2\sqrt{381}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{381}}{2*15}=\frac{-42-2\sqrt{381}}{30} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{381}}{2*15}=\frac{-42+2\sqrt{381}}{30} $
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