What are the solutions to the equation 4t^2+t-3=15 ?

The solutions to the equation 4t^2+t-3=15 are t=2,t=-9/4

Find the zeros of 4t^2+t-3=15

The zeros of 4t^2+t-3=15 are t=2,t=-9/4

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4 t^{2} + t - 3 = 15

t = 2, t = - \frac{9}{4}

Decimal Notation

t = 2, t = - 2.25

Solution steps

4 t^{2} + t - 3 = 15

Move

15

to the left side

4 t^{2} + t - 18 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 4 ( - 18 )}{2 \cdot 4}

1^{2} - 4 \cdot 4 (- 18) = 17

t_{1, 2} = \frac{- 1 \pm 17}{2 \cdot 4}

Separate the solutions

t_{1} = \frac{- 1 + 17}{2 \cdot 4}, t_{2} = \frac{- 1 - 17}{2 \cdot 4}

t = \frac{- 1 + 17}{2 \cdot 4} : 2

t = \frac{- 1 - 17}{2 \cdot 4} : - \frac{9}{4}

The solutions to the quadratic equation are:

t = 2, t = - \frac{9}{4}

z, x^{3} y^{2} z = z^{2} + x^{2} - y^{3}

x, T (x) = - \frac{1}{10} (x - 12)^{2} + 10

c^{2} + 9 c + 12 = - 4

2 x (x + 3) = 6 x + 4

22 p^{2} + 16 p - 5 = 6 p^{2}

What are the solutions to the equation 4t^2+t-3=15 ?
The solutions to the equation 4t^2+t-3=15 are t=2,t=-9/4
Find the zeros of 4t^2+t-3=15
The zeros of 4t^2+t-3=15 are t=2,t=-9/4