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10x^2+3=95
We move all terms to the left:
10x^2+3-(95)=0
We add all the numbers together, and all the variables
10x^2-92=0
a = 10; b = 0; c = -92;
Δ = b2-4ac
Δ = 02-4·10·(-92)
Δ = 3680
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3680}=\sqrt{16*230}=\sqrt{16}*\sqrt{230}=4\sqrt{230}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{230}}{2*10}=\frac{0-4\sqrt{230}}{20} =-\frac{4\sqrt{230}}{20} =-\frac{\sqrt{230}}{5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{230}}{2*10}=\frac{0+4\sqrt{230}}{20} =\frac{4\sqrt{230}}{20} =\frac{\sqrt{230}}{5} $
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