What are the solutions to the equation 5x^2-7x-13=-3 ?

The solutions to the equation 5x^2-7x-13=-3 are x=(7+sqrt(249))/(10),x=(7-sqrt(249))/(10)

Find the zeros of 5x^2-7x-13=-3

The zeros of 5x^2-7x-13=-3 are x=(7+sqrt(249))/(10),x=(7-sqrt(249))/(10)

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5 x^{2} - 7 x - 13 = - 3

x = \frac{7 + 249}{10}, x = \frac{7 - 249}{10}

Decimal Notation

x = 2.27797 \dots, x = - 0.87797 \dots

Solution steps

5 x^{2} - 7 x - 13 = - 3

Move

3

to the left side

5 x^{2} - 7 x - 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 5 ( - 10 )}{2 \cdot 5}

(- 7)^{2} - 4 \cdot 5 (- 10) = 249

x_{1, 2} = \frac{- ( - 7 ) \pm 249}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 249}{2 \cdot 5}, x_{2} = \frac{- ( - 7 ) - 249}{2 \cdot 5}

x = \frac{- ( - 7 ) + 249}{2 \cdot 5} : \frac{7 + 249}{10}

x = \frac{- ( - 7 ) - 249}{2 \cdot 5} : \frac{7 - 249}{10}

The solutions to the quadratic equation are:

x = \frac{7 + 249}{10}, x = \frac{7 - 249}{10}

x^{2} + 7 x + 10 \geq 0

x - 1 > 6

80 = (x - 6) (x - 6) (3)

(m + 4)^{2} = 0

3 x^{2} - 11 x - 14 = 0

What are the solutions to the equation 5x^2-7x-13=-3 ?
The solutions to the equation 5x^2-7x-13=-3 are x=(7+sqrt(249))/(10),x=(7-sqrt(249))/(10)
Find the zeros of 5x^2-7x-13=-3
The zeros of 5x^2-7x-13=-3 are x=(7+sqrt(249))/(10),x=(7-sqrt(249))/(10)