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

The solutions to the equation -4x^2+5x+3=0 are x=-(-5+sqrt(73))/8 ,x=(5+sqrt(73))/8

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

The zeros of -4x^2+5x+3=0 are x=-(-5+sqrt(73))/8 ,x=(5+sqrt(73))/8

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

x = - \frac{- 5 + 73}{8}, x = \frac{5 + 73}{8}

Decimal Notation

x = - 0.44300 \dots, x = 1.69300 \dots

Solution steps

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

Solve with the quadratic formula

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

5^{2} - 4 (- 4) \cdot 3 = 73

x_{1, 2} = \frac{- 5 \pm 73}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- 5 + 73}{2 ( - 4 )}, x_{2} = \frac{- 5 - 73}{2 ( - 4 )}

x = \frac{- 5 + 73}{2 ( - 4 )} : - \frac{- 5 + 73}{8}

x = \frac{- 5 - 73}{2 ( - 4 )} : \frac{5 + 73}{8}

The solutions to the quadratic equation are:

x = - \frac{- 5 + 73}{8}, x = \frac{5 + 73}{8}

(\frac{13}{2})^{2}

- x + 10 = - 3 (x + 2)

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

(- \frac{1}{2 7 - x})^{2}

- 9 \leq \frac{13}{4} x - 1 \leq - 1

What are the solutions to the equation -4x^2+5x+3=0 ?
The solutions to the equation -4x^2+5x+3=0 are x=-(-5+sqrt(73))/8 ,x=(5+sqrt(73))/8
Find the zeros of -4x^2+5x+3=0
The zeros of -4x^2+5x+3=0 are x=-(-5+sqrt(73))/8 ,x=(5+sqrt(73))/8