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charle [14.2K]
3 years ago
8

$5 PAYPAL + BRAINLIEST TO ANSWER THS

Mathematics
1 answer:
expeople1 [14]3 years ago
5 0

Answer:

Quadratic Equation:

3x^2=2x +5

\text{Standard Form: } 3x^2-2x-5=0

From the standard form of a Quadratic Function, we get:

a=3,\:b=-2,\:c=-5

Discriminant:

\Delta=\left(-2\right)^2-4\cdot \:3\left(-5\right)

\Delta=\left(-2\right)^2+4\cdot \:3\cdot \:5

\Delta=64

From the discriminant, we conclude that the equation will have two real solutions.

State that:

b^2-4ac

b^2-4ac =0:\text{The equation has 1 real solution}

b^2-4ac >0:\text{The equation has 2 real solutions}

By the way, solving the equation given:

$x=\frac{2\pm\sqrt{64}}{2\cdot \:3}$

$x=\frac{2\pm\sqrt{64}}{6}$

$x=\frac{2\pm8}{6}$

$x_{1} =\frac{10}{6}=\frac{5}{3}  $

$x_{2}=\frac{-6}{6} =-1$

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