$(a+b)^2=a^2+2ab+b^2\\------------------\\\\x^2+8x-3=0\ \ \ |add\ 3\ to\ both\ sides\\\\x^2+2x\cdot4=3\ \ \ |add\ 4^2\ to\ both\ sides\\\\x^2+2x\cdot4+4^2=3+4^2\\\\(x+4)^2=3+16\\\\(x+4)^2=19\iff x+4=-\sqrt{19}\ or\ x+4=\sqrt{19}\\|subtract\ 4\ from\ both\ sides\\\\\boxed{x=-4-\sqrt{19}\ or\ x=-4+\sqrt{19}}$

3 0

3 years ago

Read 2 more answers

Give the area if the rectangle

strojnjashka [21]

All you have to do is just multiply 6 in. and 20 in. because you are just finding the area of the rectangle not the triangle that is in the rectangle.

6in.*20in.=120in

Amswer120in

6 0

3 years ago

Which is NOT true regarding the hypotenuse? *

Usimov [2.4K]

Answer:

It helps form the right angle = FALSE

Step-by-step explanation:

The legs form the angle

6 0

2 years ago