What are the solutions to the equation (r-35)(6r+16)=0 ?

The solutions to the equation (r-35)(6r+16)=0 are r=35,r=-8/3

Find the zeros of (r-35)(6r+16)=0

The zeros of (r-35)(6r+16)=0 are r=35,r=-8/3

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(r - 35) (6 r + 16) = 0

r = 35, r = - \frac{8}{3}

Decimal Notation

r = 35, r = - 2.66666 \dots

Solution steps

(r - 35) (6 r + 16) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

r - 35 = 0 or 6 r + 16 = 0

Solve

r - 35 = 0 : r = 35

Solve

6 r + 16 = 0 : r = - \frac{8}{3}

The solutions to the quadratic equation are:

r = 35, r = - \frac{8}{3}

- a^{3} + 11 a^{2} - 39 a + 45

\frac{2}{x + 1} \leq \frac{1}{x - 2}

(x - 2)^{2} \geq 0

(k - 2) (k - 1) = - 1

(- 33)^{2}

What are the solutions to the equation (r-35)(6r+16)=0 ?
The solutions to the equation (r-35)(6r+16)=0 are r=35,r=-8/3
Find the zeros of (r-35)(6r+16)=0
The zeros of (r-35)(6r+16)=0 are r=35,r=-8/3