Question

1) х² + 12 х – 64 = 0
2) х² -11 х + 30 = 0
3) х² + 14 х + 24 = 0
4) 2х² + 3 = 3 - 7х
5) ( 3х – 1) (х + 2) = 20
РЕБЯЯТ УМОЛЯЮ КИДАЮ ВСЕ БАЛЛЫ

Answer 1

Ответ:

Объяснение:

1) $x^{2} + 12x - 64=0$

$D= 12^{2} - 4*1*(-64)= 400$

$x_{1} =\frac{-12+20 }{2}=4$

$x_{2} = \frac{-12-20}{2} = -16$

2) $x^{2} -11x+30=0$

$D= 11^{2} - 4*1*30=1$

$x_{1} = \frac{11+1}{2} =6\\x_{2} = \frac{11-1}{2} = 5$

3)

$x^{2} +14x +24=0\\D= 14^{2} - 4*1*24=100\\x_{1} = \frac{-14+10}{2} =-2\\x_{2} = \frac{-14-10}{2} =-12$

4) $2x^{2} +3=3-7x$

$2x^{2} =-7x\\2x^{2} +7x=0\\x*(2x+7)=0\\x_{1} =0 \\ x_{2} =\frac{-7}{2}$

5)$(3x-1)*(x+2)=20$

$3x^{2} +6x-x-2=20\\3x^{2} + 5x-2=20\\3x^{2} +5x-2-20=0\\3x^{2} +5x-22=0\\D=5^{2} -4*3*(-22)=289\\x_{1} = \frac{-5+17}{6} =2\\x_{2}= \frac{-5-17}{6} =\frac{-22}{6} = \frac{-11}{3}$