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(5x^2+6x-4)=0
We get rid of parentheses
5x^2+6x-4=0
a = 5; b = 6; c = -4;
Δ = b2-4ac
Δ = 62-4·5·(-4)
Δ = 116
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{116}=\sqrt{4*29}=\sqrt{4}*\sqrt{29}=2\sqrt{29}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{29}}{2*5}=\frac{-6-2\sqrt{29}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{29}}{2*5}=\frac{-6+2\sqrt{29}}{10} $
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