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(x)=-4x^2-3x+2.
We move all terms to the left:
(x)-(-4x^2-3x+2.)=0
We get rid of parentheses
4x^2+3x+x-2.=0
We add all the numbers together, and all the variables
4x^2+4x-2=0
a = 4; b = 4; c = -2;
Δ = b2-4ac
Δ = 42-4·4·(-2)
Δ = 48
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{3}}{2*4}=\frac{-4-4\sqrt{3}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{3}}{2*4}=\frac{-4+4\sqrt{3}}{8} $
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