What are the solutions to the equation x=10x^2-11x-35 ?

The solutions to the equation x=10x^2-11x-35 are x=(6+sqrt(386))/(10),x=(6-sqrt(386))/(10)

Find the zeros of x=10x^2-11x-35

The zeros of x=10x^2-11x-35 are x=(6+sqrt(386))/(10),x=(6-sqrt(386))/(10)

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x = 10 x^{2} - 11 x - 35

x = \frac{6 + 386}{10}, x = \frac{6 - 386}{10}

Decimal Notation

x = 2.56468 \dots, x = - 1.36468 \dots

Solution steps

x = 10 x^{2} - 11 x - 35

Switch sides

10 x^{2} - 11 x - 35 = x

Move

x

to the left side

10 x^{2} - 12 x - 35 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 10 ( - 35 )}{2 \cdot 10}

(- 12)^{2} - 4 \cdot 10 (- 35) = 2386

x_{1, 2} = \frac{- ( - 12 ) \pm 2 386}{2 \cdot 10}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 2 386}{2 \cdot 10}, x_{2} = \frac{- ( - 12 ) - 2 386}{2 \cdot 10}

x = \frac{- ( - 12 ) + 2 386}{2 \cdot 10} : \frac{6 + 386}{10}

x = \frac{- ( - 12 ) - 2 386}{2 \cdot 10} : \frac{6 - 386}{10}

The solutions to the quadratic equation are:

x = \frac{6 + 386}{10}, x = \frac{6 - 386}{10}

(k + 3) (2 x)^{2} + (j - x) x + m = 0

y^{2} - 15 y + 14 = - 5 y

x^{2} + 2 x + 2 = 0

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

x^{2} + \frac{4}{5} x

What are the solutions to the equation x=10x^2-11x-35 ?
The solutions to the equation x=10x^2-11x-35 are x=(6+sqrt(386))/(10),x=(6-sqrt(386))/(10)
Find the zeros of x=10x^2-11x-35
The zeros of x=10x^2-11x-35 are x=(6+sqrt(386))/(10),x=(6-sqrt(386))/(10)