What are the solutions to the equation 4x^2+48x-81=0 ?

The solutions to the equation 4x^2+48x-81=0 are x= 3/2 ,x=-27/2

Find the zeros of 4x^2+48x-81=0

The zeros of 4x^2+48x-81=0 are x= 3/2 ,x=-27/2

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4 x^{2} + 48 x - 81 = 0

x = \frac{3}{2}, x = - \frac{27}{2}

Decimal Notation

x = 1.5, x = - 13.5

Solution steps

4 x^{2} + 48 x - 81 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 48 \pm 4 8 ^{2} - 4 \cdot 4 ( - 81 )}{2 \cdot 4}

4 8^{2} - 4 \cdot 4 (- 81) = 60

x_{1, 2} = \frac{- 48 \pm 60}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 48 + 60}{2 \cdot 4}, x_{2} = \frac{- 48 - 60}{2 \cdot 4}

x = \frac{- 48 + 60}{2 \cdot 4} : \frac{3}{2}

x = \frac{- 48 - 60}{2 \cdot 4} : - \frac{27}{2}

The solutions to the quadratic equation are:

x = \frac{3}{2}, x = - \frac{27}{2}

1^{2} = m^{2} + (- \frac{3}{2})^{2}

\frac{( x - 3 ) ^{2}}{25} + \frac{( 7 - 4 ) ^{2}}{169} = 0

s^{2} + 2 s - 2 = 0

0 = x^{2} + 6 x

x^{2} + x - 4 = 0

What are the solutions to the equation 4x^2+48x-81=0 ?
The solutions to the equation 4x^2+48x-81=0 are x= 3/2 ,x=-27/2
Find the zeros of 4x^2+48x-81=0
The zeros of 4x^2+48x-81=0 are x= 3/2 ,x=-27/2