What are the solutions to the equation 4x^2+168x-540=0 ?

The solutions to the equation 4x^2+168x-540=0 are x=3,x=-45

Find the zeros of 4x^2+168x-540=0

The zeros of 4x^2+168x-540=0 are x=3,x=-45

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4 x^{2} + 168 x - 540 = 0

x = 3, x = - 45

Solution steps

4 x^{2} + 168 x - 540 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 168 \pm 16 8 ^{2} - 4 \cdot 4 ( - 540 )}{2 \cdot 4}

16 8^{2} - 4 \cdot 4 (- 540) = 192

x_{1, 2} = \frac{- 168 \pm 192}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 168 + 192}{2 \cdot 4}, x_{2} = \frac{- 168 - 192}{2 \cdot 4}

x = \frac{- 168 + 192}{2 \cdot 4} : 3

x = \frac{- 168 - 192}{2 \cdot 4} : - 45

The solutions to the quadratic equation are:

x = 3, x = - 45

\frac{x}{6} - \frac{x}{10} = \frac{1}{6}

4 x \frac{3}{2} = 32

f = \frac{1}{9} 2^{2}

\frac{4}{8} + \frac{5}{6}

4 x + \frac{1}{5} = \frac{2}{3}

What are the solutions to the equation 4x^2+168x-540=0 ?
The solutions to the equation 4x^2+168x-540=0 are x=3,x=-45
Find the zeros of 4x^2+168x-540=0
The zeros of 4x^2+168x-540=0 are x=3,x=-45