How do I use the quadratic formula to solve $4 x^{2} + x - 1 = 0$ $4x^2+x-1=0$ ?

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Answer 1 · 2014-09-30T03:46:55

$a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

$4 x^{2} + 1 x - 1 = 0$ $4x^2+1x-1=0$

$a = 4, b = 1, c = - 1$ $a=4, b=1, c=-1$

Quadratic formula

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

$x = \frac{- (1) \pm \sqrt{{(1)}^{2} - 4 (4) (- 1)}}{2 (4)}$ $x=(-(1)+-sqrt((1)^2-4(4)(-1)))/(2(4))$

$x = \frac{- 1 \pm \sqrt{1 + 16}}{8}$ $x=(-1+-sqrt(1+16))/8$

$x = \frac{- 1 \pm \sqrt{17}}{8}$ $x=(-1+-sqrt(17))/8$

How do I use the quadratic formula to solve 4x2+x−1=04x^2+x-1=0?