What are the solutions to the equation 4x^2-3x-5=0 ?

The solutions to the equation 4x^2-3x-5=0 are x=(3+sqrt(89))/8 ,x=(3-sqrt(89))/8

Find the zeros of 4x^2-3x-5=0

The zeros of 4x^2-3x-5=0 are x=(3+sqrt(89))/8 ,x=(3-sqrt(89))/8

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4 x^{2} - 3 x - 5 = 0

x = \frac{3 + 89}{8}, x = \frac{3 - 89}{8}

Solution steps

4 x^{2} - 3 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 4 ( - 5 )}{2 \cdot 4}

(- 3)^{2} - 4 \cdot 4 (- 5) = 89

x_{1, 2} = \frac{- ( - 3 ) \pm 89}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 89}{2 \cdot 4}, x_{2} = \frac{- ( - 3 ) - 89}{2 \cdot 4}

x = \frac{- ( - 3 ) + 89}{2 \cdot 4} : \frac{3 + 89}{8}

x = \frac{- ( - 3 ) - 89}{2 \cdot 4} : \frac{3 - 89}{8}

The solutions to the quadratic equation are:

x = \frac{3 + 89}{8}, x = \frac{3 - 89}{8}

\frac{1}{3} (p + 1) \leq \frac{3}{6} (p - 2)

- (3)^{3}

2 x = 30

e^{x} x^{2} + 2 e^{x} x

48 p - 32 = 51 p - 65

What are the solutions to the equation 4x^2-3x-5=0 ?
The solutions to the equation 4x^2-3x-5=0 are x=(3+sqrt(89))/8 ,x=(3-sqrt(89))/8
Find the zeros of 4x^2-3x-5=0
The zeros of 4x^2-3x-5=0 are x=(3+sqrt(89))/8 ,x=(3-sqrt(89))/8