Question

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

Answer 1

-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) = (-b \pm √(\Delta))/(2a) \\\;\\ \\\;\\ x_(1,2) = (6\pm√(16))/(2\cdot1) = (6\pm4)/(2)`


`\it x_1= (6-4)/(2) = (2)/(2) =1 \\\;\\ \\\;\\ x_2 = (6+4)/(2) = (10)/(2) = 5`

Answer 2

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