What are the solutions to the equation 5x^2=19x+4 ?

The solutions to the equation 5x^2=19x+4 are x=4,x=-1/5

Find the zeros of 5x^2=19x+4

The zeros of 5x^2=19x+4 are x=4,x=-1/5

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5 x^{2} = 19 x + 4

x = 4, x = - \frac{1}{5}

Decimal Notation

x = 4, x = - 0.2

Solution steps

5 x^{2} = 19 x + 4

Move

4

to the left side

5 x^{2} - 4 = 19 x

Move

19 x

to the left side

5 x^{2} - 4 - 19 x = 0

Write in the standard form

a x^{2} + b x + c = 0

5 x^{2} - 19 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 19 ) \pm ( - 19 ) ^{2} - 4 \cdot 5 ( - 4 )}{2 \cdot 5}

(- 19)^{2} - 4 \cdot 5 (- 4) = 21

x_{1, 2} = \frac{- ( - 19 ) \pm 21}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 19 ) + 21}{2 \cdot 5}, x_{2} = \frac{- ( - 19 ) - 21}{2 \cdot 5}

x = \frac{- ( - 19 ) + 21}{2 \cdot 5} : 4

x = \frac{- ( - 19 ) - 21}{2 \cdot 5} : - \frac{1}{5}

The solutions to the quadratic equation are:

x = 4, x = - \frac{1}{5}

64 y^{2} = 25

- 4 x^{2} + 12 x - 9 = 0

\frac{1}{2} (- 6 w + 8) - 5 = 3 w - 1

\frac{1}{2} x^{2} = \frac{2}{5} x^{2}

(5 - 7 i) (8 + 13 i)

What are the solutions to the equation 5x^2=19x+4 ?
The solutions to the equation 5x^2=19x+4 are x=4,x=-1/5
Find the zeros of 5x^2=19x+4
The zeros of 5x^2=19x+4 are x=4,x=-1/5