How do you solve x^2-5x-1 = 0 using the quadratic formula?

Apr 14, 2016

${x}_{1} = 0 , 195$
${x}_{2} = 5 , 195$

Explanation:

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

${x}_{\text{1,2"=(-b±sqrt(b^2-4*a*c))/(2*a)" the quadratic formula}}$

${x}^{2} - 5 x - 1 = 0$

$a = 1 \text{ "b=-5" } c = - 1$

$\Delta = \sqrt{{b}^{2} - 4 \cdot a \cdot c}$

$\Delta = \sqrt{25 + 4 \cdot 1 \cdot 1}$

Delta=±5,39

${x}_{1} = \frac{- b - \Delta}{2 \cdot a}$

${x}_{1} = \frac{5 - 5 , 39}{2 \cdot 1}$

${x}_{1} = \frac{0 , 39}{2}$

${x}_{1} = 0 , 195$

${x}_{2} = \frac{- b + \Delta}{2 \cdot a}$

${x}_{2} = \frac{5 + 5 , 39}{2}$

${x}_{2} = \frac{10 , 39}{2}$

${x}_{2} = 5 , 195$