Question

solve the equation $-2x^2 - 13x + 20 = -3x^2$ to the nearest tenth.

Explanation:

Step1: Simplify the equation

First, we want to get all terms on one side of the equation. Add \(3x^{2}\) to both sides of the equation \(-2x^{2}-13x + 20=-3x^{2}\).

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Step2: Use the quadratic formula

For a quadratic equation \(ax^{2}+bx + c = 0\) (here \(a = 1\), \(b=-13\), \(c = 20\)), the quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac\). Substitute \(a = 1\), \(b=-13\), \(c = 20\) into the discriminant formula:

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Then, substitute \(a = 1\), \(b=-13\), \(\Delta = 89\) into the quadratic formula:

$$ x=\frac{13\pm\sqrt{89}}{2} $$

Step3: Calculate the two solutions

We know that \(\sqrt{89}\approx9.43\).
For the plus sign:

$$ x_{1}=\frac{13 + 9.43}{2}=\frac{22.43}{2}=11.215\approx11.2 $$

For the minus sign:

$$ x_{2}=\frac{13-9.43}{2}=\frac{3.57}{2}=1.785\approx1.8 $$

Answer:

The solutions of the equation are \(x\approx1.8\) and \(x\approx11.2\)