$x^2+2x-4=0\qquad\text{add 4 to both sides}\\\\x^2+2x=4\\\\x^2+2(x)(1)=4\qquad\text{add}\ 1^2=1\ \text{to both sides}\\\\\underbrace{x^2+2(x)(1)+1^2}=4+1^2\qquad\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\(x+1)^2=5\to x+1=\pm\sqrt5\qquad\text{subtract 1 from both sides}\\\\x=-1\pm\sqrt5$

4 0

3 years ago

Help on math plz <br>15 and 16 explain

Anit [1.1K]

For 15:
equate both left-hand sides and you will get a quadratic ..

Solve it for x, you get:

x=4, x=-1.75

B

For 16:
the vertex would be (h,k) = (2,4)

A

6 0

3 years ago

Please answer correctly !!!!!!!!!!!!!! Will mark Brianliest !!!!!!!!!!!!!!!!!

Maksim231197 [3]

........................

4 0

3 years ago

Read 2 more answers