What are the solutions to the equation 2r^2+5r-3=0 ?

The solutions to the equation 2r^2+5r-3=0 are r= 1/2 ,r=-3

Find the zeros of 2r^2+5r-3=0

The zeros of 2r^2+5r-3=0 are r= 1/2 ,r=-3

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2 r^{2} + 5 r - 3 = 0

r = \frac{1}{2}, r = - 3

Decimal Notation

r = 0.5, r = - 3

Solution steps

2 r^{2} + 5 r - 3 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

5^{2} - 4 \cdot 2 (- 3) = 7

r_{1, 2} = \frac{- 5 \pm 7}{2 \cdot 2}

Separate the solutions

r_{1} = \frac{- 5 + 7}{2 \cdot 2}, r_{2} = \frac{- 5 - 7}{2 \cdot 2}

r = \frac{- 5 + 7}{2 \cdot 2} : \frac{1}{2}

r = \frac{- 5 - 7}{2 \cdot 2} : - 3

The solutions to the quadratic equation are:

r = \frac{1}{2}, r = - 3

(7 x - 3 y) (2 x - 3 y) - (3 y - 2 x)^{2}

8 x^{- \frac{1}{2}} + 6 x^{\frac{1}{2}} + x^{\frac{3}{2}}

\frac{n}{6.5} = 16.6

- 1 < \frac{n}{2} + 1

What are the solutions to the equation 2r^2+5r-3=0 ?
The solutions to the equation 2r^2+5r-3=0 are r= 1/2 ,r=-3
Find the zeros of 2r^2+5r-3=0
The zeros of 2r^2+5r-3=0 are r= 1/2 ,r=-3