How do you solve 3x^2+5x=23x2+5x=2 by using the quadratic formula?

1 Answer
Lucio Falabella
Oct 6, 2016

x_1=-2x1=−2
x_2=1/3x2=13

Explanation:

Given a quadratic equation

ax^2+bx+c=0ax2+bx+c=0

the Quadratic Formula tell us that:

x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)x1,2=−b±√b2−4ac2a

You need to lead your equation in the canonical form:

3x^2+5x=2=>3x^2+5x-2=03x2+5x=2⇒3x2+5x−2=0

Therefore you find:

a=3; b=5; c=-2a=3;b=5;c=−2

So you can substitute them in the quadratic formula:

x_(1,2)=(-5+-sqrt(5^2-4*3*(-2)))/(2*3)=(-5+-sqrt(25+24))/6=x1,2=−5±√52−4⋅3⋅(−2)2⋅3=−5±√25+246=
=(-5+-sqrt(49))/6=(-5+-7)/6=−5±√496=−5±76

x_1=(-5-7)/6=-cancel(12)^2/cancel(6)_1=-2
x_2=(-5+7)/6=cancel(2)^1/cancel(6)_3=1/3

graph{3x^2+5x-2 [-2.434, 2.433, -1.214, 1.218]}

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