"How do you solve 7(x2−2x+3)=−3(x2−5x)?" was answered by our experts

High School
Algebra
Algebra I
Quadratic function and equation

Rylan Duncan

Answered question

2022-01-22

How do you solve

7 (x^{2} - 2 x + 3) = - 3 (x^{2} - 5 x)

Answer & Explanation

Howard Gallagher

Beginner2022-01-23Added 13 answers

Step 1
Let's do the multiplications at both sides:
$7 x^{2} - 14 + 21 = - 3 x^{2} + 15 x$
Add $3 x^{2}$ at both sides:
$10 x^{2} - 14 x + 21 = 15 x$
Subtract 15x from both sides:
$10 x^{2} - 29 x + 21 = 0$
Step 2
So we have a quadratic equation of the form
$a x^{2} + b x + c$
where $a = 10 Z, b = - 29$ , and $c = 21$ .
Let's calculate the discriminant:
$Δ = b^{2} - 4 a c = (- 29^{2}) - 4 \times 10 \times 21 = 841 - 840 = 1$
Since $Δ > 0$ , there are two solutions $x_{1}$ and $x_{2}$ , given by
$x_{1, 2} = \frac{- b \pm \sqrt{Δ}}{2 a}$
Since $Δ = 1$ , its square root is also 1. The two solutions are thus $\frac{29 \pm 1}{20}$
So,
$x_{1} = \frac{29 + 1}{20} = \frac{3}{2}$
is the first solution, while
$x_{2} = \frac{29 - 1}{20} = \frac{7}{5}$
is the second.

izumrledk

Beginner2022-01-24Added 15 answers

Step 1

7 (x^{2} - 2 x + 3) = - 3 (x^{2} - 5 x)

7 x^{2} - 14 x + 21 = - 3 x^{2} + 15 x

10 x^{2} - 29 x + 21 = 0

x_{1, 2} = \frac{29 \pm \sqrt{{(29)}^{2} - 4 (10 \times 21)}}{2 \times 10}

= \frac{29 \pm \sqrt{841 - 840}}{20} = \frac{29 \pm 1}{20}

x_{1} = \frac{30}{20} = \frac{3}{2}

x_{2} = \frac{28}{20} = \frac{14}{20} = \frac{7}{5}

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