Question

solve the equation for all real solutions in simplest form.\\(2r^{2} + 3r - 5 = -1\\)

Explanation:

Step1: Rewrite in standard quadratic form

Add 1 to both sides to set equation to 0:
$2r^2 + 3r - 5 + 1 = 0$
$2r^2 + 3r - 4 = 0$

Step2: Identify quadratic coefficients

For $ax^2+bx+c=0$, define:
$a=2$, $b=3$, $c=-4$

Step3: Apply quadratic formula

Use $r=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$:
$r=\frac{-3\pm\sqrt{3^2-4(2)(-4)}}{2(2)}$

Step4: Simplify discriminant

Calculate the value under the square root:
$\sqrt{9 + 32} = \sqrt{41}$

Step5: Finalize solution

Substitute discriminant back into formula:
$r=\frac{-3\pm\sqrt{41}}{4}$

Answer:

$r=\frac{-3+\sqrt{41}}{4}$ and $r=\frac{-3-\sqrt{41}}{4}$