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42x^2-15=0
a = 42; b = 0; c = -15;
Δ = b2-4ac
Δ = 02-4·42·(-15)
Δ = 2520
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2520}=\sqrt{36*70}=\sqrt{36}*\sqrt{70}=6\sqrt{70}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{70}}{2*42}=\frac{0-6\sqrt{70}}{84} =-\frac{6\sqrt{70}}{84} =-\frac{\sqrt{70}}{14} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{70}}{2*42}=\frac{0+6\sqrt{70}}{84} =\frac{6\sqrt{70}}{84} =\frac{\sqrt{70}}{14} $
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