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Question

2x2 +$\frac{x-1}{5}$ = 0

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Solution

Given: 2x2 +$\frac{x-1}{5}$ = 0 On rewriting this equation in standard form of the quadratic equation, we get: 10x2 + x – 1 = 0 On comparing this equation with ax2 + bx + c = 0, we get: a = 10, b = 1, c = –1 For the quadratic equation ax2 + bx + c = 0, we know that the quadratic formula is $x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$ On substituting a = 10, b = 1, c = –1, we get: $x=\frac{-1±\sqrt{{\left(1\right)}^{2}-4\left(10\right)\left(-1\right)}}{2\left(10\right)}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-1±\sqrt{1+40}}{2\left(10\right)}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-1±\sqrt{41}}{20}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-1+\sqrt{41}}{20},\frac{-1-\sqrt{41}}{20}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{the}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{equation}\text{is}x=\frac{-1+\sqrt{41}}{20},\frac{-1-\sqrt{41}}{20}$

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