Question

penyelesaian persamaan kuadrat 20x^2-13x+2=0

Answer 1

Dengan Bhaskara:


$20x^2 - 13x + 2 = 0$


$x = \dfrac{-b \pm \sqrt{b^2 -4*a*c}}{2*a}$

a=20, b=−13, c=2

Δ=b2−4ac
Δ=(−13)2−4*(20)*(2)
Δ=169−160
Δ=9

$x = \dfrac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ \\ x = \dfrac{-(-13) \pm \sqrt{9}}{2*20} \\ \\ \\ x = \dfrac{13 \pm 3}{40} \\ \\ \\ x' = \dfrac{13 + 3}{40} \\ \\ \\ x' = \dfrac{16}{40} \\ \\ \\ x' = \dfrac{2}{5} \\ \\ \\ x'' = \dfrac{13 - 3}{40} \\ \\ \\ x'' = \dfrac{10}{40} \\ \\ \\ x'' = \dfrac{1}{4}$


S = {$\dfrac{2}{5}, \ \ \dfrac{1}{4}$}