What are the solutions to the equation 5x^2-16x-253=4x+7 ?

The solutions to the equation 5x^2-16x-253=4x+7 are x=2(1+sqrt(14)),x=2(1-sqrt(14))

Find the zeros of 5x^2-16x-253=4x+7

The zeros of 5x^2-16x-253=4x+7 are x=2(1+sqrt(14)),x=2(1-sqrt(14))

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5 x^{2} - 16 x - 253 = 4 x + 7

x = 2 (1 + 14), x = 2 (1 - 14)

Decimal Notation

x = 9.48331 \dots, x = - 5.48331 \dots

Solution steps

5 x^{2} - 16 x - 253 = 4 x + 7

Move

7

to the left side

5 x^{2} - 16 x - 260 = 4 x

Move

4 x

to the left side

5 x^{2} - 20 x - 260 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 5 ( - 260 )}{2 \cdot 5}

(- 20)^{2} - 4 \cdot 5 (- 260) = 2014

x_{1, 2} = \frac{- ( - 20 ) \pm 20 14}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 20 14}{2 \cdot 5}, x_{2} = \frac{- ( - 20 ) - 20 14}{2 \cdot 5}

x = \frac{- ( - 20 ) + 20 14}{2 \cdot 5} : 2 (1 + 14)

x = \frac{- ( - 20 ) - 20 14}{2 \cdot 5} : 2 (1 - 14)

The solutions to the quadratic equation are:

x = 2 (1 + 14), x = 2 (1 - 14)

x (5 x - 6) = 11

x^{2} + 15 = - 16 x

8 x^{2} + 16 = 0

(x - 3)^{2} - 1 = 0

(x - 3)^{2} - 2 = 0

What are the solutions to the equation 5x^2-16x-253=4x+7 ?
The solutions to the equation 5x^2-16x-253=4x+7 are x=2(1+sqrt(14)),x=2(1-sqrt(14))
Find the zeros of 5x^2-16x-253=4x+7
The zeros of 5x^2-16x-253=4x+7 are x=2(1+sqrt(14)),x=2(1-sqrt(14))