Question

Anyone can help me to solve this!​

Answer 1

Solve following by using middle term factoring and then set to zero.

6.

$\sf \rightarrow x^2 + 7x+6=0$

$\sf \rightarrow x^2 +x+ 6x+6=0$

$\sf \rightarrow x(x +1)+ 6(x+1)=0$

$\sf \rightarrow (x+ 6)(x+1)=0$

$\sf \rightarrow x=-6,-1$

7.

$\sf \rightarrow x^2 - 10x+21=0$

$\sf \rightarrow x^2 - 7x-3x+21=0$

$\sf \rightarrow x(x - 7)-3(x-7)=0$

$\sf \rightarrow (x-3)(x-7)=0$

$\sf \rightarrow x = 3,7$

8.

$\sf \rightarrow x^2-5x-14=0$

$\sf \rightarrow x^2-7x+2x-14=0$

$\sf \rightarrow x(x-7)+2(x-7)=0$

$\sf \rightarrow (x+2)(x-7)=0$

$\sf \rightarrow x=-2,7$

9.

$\sf \rightarrow 2x^2+9x+4=0$

$\sf \rightarrow 2x^2+8x+x+4=0$

$\sf \rightarrow 2x(x+4)+1(x+4)=0$

$\sf \rightarrow (2x+1)(x+4)=0$

$\sf \rightarrow x=-4, -\dfrac{1}{2}$

10.

$\sf \rightarrow 3x^2-17x+10=0$

$\sf \rightarrow 3x^2-2x-15x+10=0$

$\sf \rightarrow x(3x-2)-5(3x-2)=0$

$\sf \rightarrow (x-5)(3x-2)=0$

$\sf \rightarrow x=\dfrac{2}{3}, 5$

Answer 2

Answer:

6) x= -1,-6

7) x=3,7

8) x=7,-2

9) x= -1/2, -4

10) x= 5, 2/3

The first three you can solve by factoring.  For the last two I used the quadratic formula bc its easiest.