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2x^2+5x+5/4=0
We multiply all the terms by the denominator
2x^2*4+5x*4+5=0
Wy multiply elements
8x^2+20x+5=0
a = 8; b = 20; c = +5;
Δ = b2-4ac
Δ = 202-4·8·5
Δ = 240
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{15}}{2*8}=\frac{-20-4\sqrt{15}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{15}}{2*8}=\frac{-20+4\sqrt{15}}{16} $
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