What are the solutions to the equation 4x^2+13x-35=0 ?

The solutions to the equation 4x^2+13x-35=0 are x= 7/4 ,x=-5

Find the zeros of 4x^2+13x-35=0

The zeros of 4x^2+13x-35=0 are x= 7/4 ,x=-5

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4 x^{2} + 13 x - 35 = 0

x = \frac{7}{4}, x = - 5

Decimal Notation

x = 1.75, x = - 5

Solution steps

4 x^{2} + 13 x - 35 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 4 ( - 35 )}{2 \cdot 4}

1 3^{2} - 4 \cdot 4 (- 35) = 27

x_{1, 2} = \frac{- 13 \pm 27}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 13 + 27}{2 \cdot 4}, x_{2} = \frac{- 13 - 27}{2 \cdot 4}

x = \frac{- 13 + 27}{2 \cdot 4} : \frac{7}{4}

x = \frac{- 13 - 27}{2 \cdot 4} : - 5

The solutions to the quadratic equation are:

x = \frac{7}{4}, x = - 5

x^{2} + 11 x - 2

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

(2 x + 5) (x - 1) = - 6

u^{2} = - 25

12 x^{2} - 30 x + 12 = 0

What are the solutions to the equation 4x^2+13x-35=0 ?
The solutions to the equation 4x^2+13x-35=0 are x= 7/4 ,x=-5
Find the zeros of 4x^2+13x-35=0
The zeros of 4x^2+13x-35=0 are x= 7/4 ,x=-5