Question

Решите квадратные уравнения: 5х^2+3х-8=0 х^2-4х+3=0

Answer 1

$5x^2+3x-8=0 \\ D=b^2-4ac = 3^2-4*5*(-8) = 9 + 160 = \sqrt{169} = 13 \\ x1= \frac{-b+ \sqrt{D} }{2a} = \frac{-3+13}{10} =1 \\ x2 = \frac{-b+ \sqrt{D} }{2a} = \frac{-3-13}{10} = \frac{-16}{10} =-1.6$

$x^2-4x+3=0 \\ D=b^2-4ac = (-4)^2-4*1*3 = 16-12 = \sqrt{4} = 2 \\ x1= \frac{-b+ \sqrt{D} }{2a} = \frac{4+2}{2} =3 \\ x2 = \frac{-b+ \sqrt{D} }{2a} = \frac{4-2}{2} = 1$

Answer 2

5х^2 + 3х - 8 = 0
а = 5 b = 3 с = -8
Д = 9 - 4 • 5 • ( - 8 ) = 9 + 160 = 169 > 0. 13.
х1 = -3 + 13 / 10 = 10/10 = 1.
х2 = -3 - 13 / 10 = -16/10 = -1,6.
ответ : х = -1,6 ; 1.

х^2 - 4х + 3 = 0
а = 1 b = -4 с = 3
Д = 16 - 4 • 1 • 3 = 16 - 12 = 4 > 0. 2.
х1 = 4 + 2 / 2 = 6/2 = 3.
х2 = 4 - 2 / 2 = 2/2 = 1.
ответ : х = 1 ; 3.
Удачи)))