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5x^2+25x-34=0
a = 5; b = 25; c = -34;
Δ = b2-4ac
Δ = 252-4·5·(-34)
Δ = 1305
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1305}=\sqrt{9*145}=\sqrt{9}*\sqrt{145}=3\sqrt{145}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-3\sqrt{145}}{2*5}=\frac{-25-3\sqrt{145}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+3\sqrt{145}}{2*5}=\frac{-25+3\sqrt{145}}{10} $
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