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(x)=4x^2+x-5
We move all terms to the left:
(x)-(4x^2+x-5)=0
We get rid of parentheses
-4x^2+x-x+5=0
We add all the numbers together, and all the variables
-4x^2+5=0
a = -4; b = 0; c = +5;
Δ = b2-4ac
Δ = 02-4·(-4)·5
Δ = 80
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{5}}{2*-4}=\frac{0-4\sqrt{5}}{-8} =-\frac{4\sqrt{5}}{-8} =-\frac{\sqrt{5}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{5}}{2*-4}=\frac{0+4\sqrt{5}}{-8} =\frac{4\sqrt{5}}{-8} =\frac{\sqrt{5}}{-2} $
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