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aivan3 [116]
4 years ago
11

Solve the following quadratic equation using the quadratic formula.

Mathematics
2 answers:
IRINA_888 [86]4 years ago
8 0

x=4-3i/5 x=4+3-/5

plato users!!!!

sdas [7]4 years ago
6 0
ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a};\ x_2=\dfrac{-b+\sqrt\Delta}{2a}

<span>5x^2-8x+5=0\to a=5;\ b=-8;\ c=5\\\\\Delta=(-8)^2-4\cdot5\cdot5=64-100=-36\\\\\sqrt\Delta=\sqrt{-36}=6i

x_1=\dfrac{-(-8)-6i}{2\cdot5}=\dfrac{8-6i}{10}=\dfrac{4-3i}{5}\\\\x_2=\dfrac{-(-8)+6i}{2\cdot5}=\dfrac{8+6i}{10}=\dfrac{4+3i}{5}
</span>
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Answer:

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Step-by-step explanation:

The <em>incenter</em> is the center of the largest circle that can be inscribed in the triangle. That circle is called the <em>incircle</em>. The incenter is at the point of intersection of the angle bisectors. For a right triangle, the <em>inradius</em> (the radius of the incircle), is found from a simple formula:

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<h3>What is a proportional relationship?</h3>

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