math equation helpppppp?

Question

solve and give answer to nearest tenth
x^2 + x + 1 = 5

also, find the x intercepts of
y = sin(2x) where 0 degrees<=x<=180 degrees

thanks for the help! :)

Answer 1

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

x = [-1 +/- (1 + 16)^1/2 ]/ 2

x = [-1 +/- (17)^1/2]/ 2

2.) y = sin(2x)
Set Y = 0 to find the X-intercepts.
sin(2x) = 0
2sinx cosx = 0

1st case,
sinx = 0
x = 0, 180

2nd case,
cosx = 0
x = 90

x = 0, 90, 180 degrees are the X-intercepts.

Answer 2

Question 1
x² + x - 4 = 0
x = [- 1 ± √(1 + 16) ] / 2
x = [- 1 ± √17] / 2
x = 1.6 , x = - 2.6

Question 2
x intercepts are 0° , 90° , 180° for x ≤ 180°

Answer 3

x^2 + x + 1 = 5
x^2 + x + 1 - 5 = 0
x^2+x-4=0
for ax^2+bx+c=0
x= {-b±√(b^2-4ac)}/2a

Therefore,
x={-1±√(1+16)}/2
x=(-1+√17)/2 & (-1-√17)/2
-------------------------------
for y-intercept, x=0
y=sin(2x) = sin(0)
y=0.

The graph passes through (0,0). Its y-intercept is 0.