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5x^2+10x-142=0
a = 5; b = 10; c = -142;
Δ = b2-4ac
Δ = 102-4·5·(-142)
Δ = 2940
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2940}=\sqrt{196*15}=\sqrt{196}*\sqrt{15}=14\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-14\sqrt{15}}{2*5}=\frac{-10-14\sqrt{15}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+14\sqrt{15}}{2*5}=\frac{-10+14\sqrt{15}}{10} $
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