Question

Need help please and thank you. 1. Solve. 8x^2+5=35. Round to the nearest hundredth. Enter your answers in the boxes. The solutions are approximately and . 2. Solve. 0=x^2+14x+40 Enter your answers in the boxes. The solutions are and .

Answer 1

Answer: 1.94 and -1.94 (just took the test, got 100%)

Answer 2

Part 1) The solutions are approximately
`x=1.94` and
`x=-1.94`

Part 2) The solutions are
`x=-4` and
`x=-10`

Explanation:

Part 1)

in this problem we have

`8x^(2) +5=35`

`8x^(2)=35-5`

`8x^(2)=30`

`x^(2)=30/8`

square root both sides

`x=(+/-)√((30/8))\\ \\x=(+/-)1.94`

The solutions are
`x=1.94` and
`x=-1.94`

Part 2)

we know that

The formula to solve a quadratic equation of the form
`ax^(2) +bx+c=0` is equal to

`x=\frac{-b(+/-)\sqrt{b^(2)-4ac}} {2a}`

in this problem we have

`0=x^(2) +14x+40`

so

`a=1\\b=14\\c=40`

substitute in the formula

`x=\frac{-14(+/-)\sqrt{14^(2)-4(1)(40)}} {2(1)}`

`x=\frac{-14(+/-)√(36)} {2}`

`x=\frac{-14(+/-)6} {2}`

`x=\frac{-14(+)6} {2}=-4`

`x=\frac{-14(-)6} {2}=-10`

The solutions are
`x=-4` and
`x=-10`