What are the solutions to the equation 2x^2+10x-31=7x ?

The solutions to the equation 2x^2+10x-31=7x are x=(-3+sqrt(257))/4 ,x=(-3-sqrt(257))/4

Find the zeros of 2x^2+10x-31=7x

The zeros of 2x^2+10x-31=7x are x=(-3+sqrt(257))/4 ,x=(-3-sqrt(257))/4

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2 x^{2} + 10 x - 31 = 7 x

x = \frac{- 3 + 257}{4}, x = \frac{- 3 - 257}{4}

Decimal

x = 3.25780 \dots, x = - 4.75780 \dots

Solution steps

2 x^{2} + 10 x - 31 = 7 x

Move

7 x

to the left side

2 x^{2} + 3 x - 31 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 ( - 31 )}{2 \cdot 2}

3^{2} - 4 \cdot 2 (- 31) = 257

x_{1, 2} = \frac{- 3 \pm 257}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 3 + 257}{2 \cdot 2}, x_{2} = \frac{- 3 - 257}{2 \cdot 2}

x = \frac{- 3 + 257}{2 \cdot 2} : \frac{- 3 + 257}{4}

x = \frac{- 3 - 257}{2 \cdot 2} : \frac{- 3 - 257}{4}

The solutions to the quadratic equation are:

x = \frac{- 3 + 257}{4}, x = \frac{- 3 - 257}{4}

x = - 1

8.99 \cdot 1 0^{9} \cdot \frac{1 \cdot 1 0 ^{- 6}}{3 ^{2}}

2 - 3 x = 11

\frac{7 x - 1}{3} - \frac{5 - 2 x}{2} = \frac{4 x - 3}{4} + \frac{1 + 4 x}{3}

\frac{( - 2 ) ^{8} \cdot 5 ^{3}}{5 ^{4} \cdot 2 ^{10} \cdot 10}

What are the solutions to the equation 2x^2+10x-31=7x ?
The solutions to the equation 2x^2+10x-31=7x are x=(-3+sqrt(257))/4 ,x=(-3-sqrt(257))/4
Find the zeros of 2x^2+10x-31=7x
The zeros of 2x^2+10x-31=7x are x=(-3+sqrt(257))/4 ,x=(-3-sqrt(257))/4