Question

use the quadratic formula to solve the equation -x^2+6x-5=0

Answer 1

a=-1 b=6 c= -5

x1 = [-(6) + sqrt ( 6^2 -4*(-1)*(-5))] / 2*(-1)

=[-6+sqrt (16)] / (-2)
=[-6+4]/-2
= 1

x2= [-(6) - sqrt ( 6^2 -4*(-1)*(-5))] / 2*(-1)

=[-6-sqrt (16)] / (-2)
=[-6-4]/-2
= 5

x1 = 1
x2 = 5

Answer 2

-x^2 + 6x - 5 = 0 |·(-1) ⇔ x² -6x + 5 = 0, a = 1,  b = -6,  c = 5

Δ = b² - 4ac

Δ = (-6)² - 4·1·5 = 36 - 20 = 16

$\it x_{1,2} = \dfrac{-b \pm \sqrt{\Delta}}{2a} \\\;\\ \\\;\\ x_{1,2} = \dfrac{6\pm\sqrt{16}}{2\cdot1} = \dfrac{6\pm4}{2}$

$\it x_1= \dfrac{6-4}{2} = \dfrac{2}{2} =1 \\\;\\ \\\;\\ x_2 = \dfrac{6+4}{2} = \dfrac{10}{2} = 5$