What are the solutions to the equation (x^2-(17x))/(4+1)=0 ?

The solutions to the equation (x^2-(17x))/(4+1)=0 are x=17,x=0

Find the zeros of (x^2-(17x))/(4+1)=0

The zeros of (x^2-(17x))/(4+1)=0 are x=17,x=0

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\frac{x ^{2} - ( 17 x )}{4 + 1} = 0

x = 17, x = 0

Solution steps

\frac{x ^{2} - ( 17 x )}{4 + 1} = 0

Simplify

4 + 1 : 5

\frac{x ^{2} - ( 17 x )}{5} = 0

Multiply both sides by

5

x^{2} - 17 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 17 ) \pm ( - 17 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 17)^{2} - 4 \cdot 1 \cdot 0 = 17

x_{1, 2} = \frac{- ( - 17 ) \pm 17}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 17 ) + 17}{2 \cdot 1}, x_{2} = \frac{- ( - 17 ) - 17}{2 \cdot 1}

x = \frac{- ( - 17 ) + 17}{2 \cdot 1} : 17

x = \frac{- ( - 17 ) - 17}{2 \cdot 1} : 0

The solutions to the quadratic equation are:

x = 17, x = 0

2 p^{2} + k^{3} = 15

(x) = - 5 (x + 20) \cdot (x - 80)

x, x^{2} + y^{2} + x y = 192

(16 x)^{2} + (9 x)^{2} = 4 3^{2}

x, ln (y) - \frac{1}{y ^{2}} = \frac{x ^{2}}{2} + c

What are the solutions to the equation (x^2-(17x))/(4+1)=0 ?
The solutions to the equation (x^2-(17x))/(4+1)=0 are x=17,x=0
Find the zeros of (x^2-(17x))/(4+1)=0
The zeros of (x^2-(17x))/(4+1)=0 are x=17,x=0