Question

Solve the following quadratic equations.... ​

Answer 1

Answer:

Step-by-step explanation:

x2-3x-10=0

x2+2x-5x-10=0-----------------[+2x-5x=3x]

x[x+2]-5[x+2]=0

[x+2][x-5]=0

x+2=0 or x-5=0

x=-2 or x=5

2]2x2+x-6=0

2x2+4x-3x-6=0---------[4x-3x=x]

2x[x+2]-3[x+2]=0

[2x-3][x+2]=0

2x-3=0 or x+2=0

2x=3 or x=-2

x=3/2 or x=-2

3]2x2-7x+3=0

2x2-6x-x+3=0----------[6x-x=7x]

2x[x-3]-1[x-3]

[2x-1][x-3]=0

2x-1=0 or x-3=0

2x=1 or x=3

x=1/2 or x=3

4]i guess it's polynomial

Answer 2

1)

x^2 -3x -10=0.

D = b^2- 4ac = 9+40= 49.

Now , x =( -b-√D)/2a or , x =(-b +√D )2a

→ x = (3-7)/2= -2 or , x = (3+7)/2= 5

2)2x^2 + x -6=0

→ 2x^2 + 4x -3x -6=0

→ 2x(x+2)-3(x+2)= 0

→ (2x-3)(x+2)= 0

→x =3/2 or , -2 .

3) 2x ^2 -7x +3 = 0

→2x^2 - 6x -x +3=0

→2x (x-3)-1(x-3)= 0

→ (2x-1)(x-3)= 0

→ x =1 /2 or , 3

4)2x^2 + x -4= 0

→ D = b^2 - 4ac = 1+ 32= 33 .

Now , x =( -b -√D)/2a or , (-b+√D)/2a

→ x = (-1-√33)/2 or , x = (√33-1)/2